MH1810 Math

Chapter 4b: Limits and Continuity

Imron Rosyadi

4.9 Techniques in Finding Limits

4.9.1 Eliminating Zero Denominators

When direct substitution results in an indeterminate form like \(\frac{0}{0}\), we often need to simplify the expression by factoring and canceling common factors.

Example 4.9.1 / 4.9.2:

Evaluate \(\lim_{x \to -2}\frac{3x + 6}{x^2 - 4}\).

• Question: Can we apply the quotient limit law: \(\frac{\lim_{x \to -2} (3x+6)}{\lim_{x \to -2} (x^2-4)}\)?

  No, because the denominator \(\lim_{x \to -2} (x^2-4) = (-2)^2 - 4 = 4 - 4 = 0\). This gives \(\frac{0}{0}\).

• Solution: We simplify first: \[ \begin{aligned} \lim_{x \to -2} \frac{3x + 6}{x^2 - 4} &= \lim_{x \to -2} \frac{3(x + 2)}{(x + 2)(x - 2)} & \text{(Factor numerator and denominator)} \\ &= \lim_{x \to -2} \frac{3}{x - 2} & \text{(Cancel common factor (x+2), valid for } x \neq -2) \\ &= \frac{3}{-2 - 2} & \text{(Now substitute, as denominator is not zero)} \\ &= -\frac{3}{4}. \end{aligned} \]

4.9 Techniques in Finding Limits

Example 4.9.3:

\(\lim_{x \to -3}\frac{x^3 + 4x^2 + 4x + 3}{-x^3 - 2x^2 + 5x + 6}\)

• Direct substitution gives \(\frac{0}{0}\). Both polynomials have a factor \((x+3)\) by the Factor Theorem.

• By long division (or synthetic division):

  \(x^3 + 4x^2 + 4x + 3 = (x + 3)(x^2 + x + 1)\)

  \(-x^3 - 2x^2 + 5x + 6 = (x + 3)(-x^2 + x + 2)\)

• Now evaluate the limit: \[ \lim_{x \to -3} \frac{(x + 3)(x^2 + x + 1)}{(x + 3)(-x^2 + x + 2)} = \lim_{x \to -3} \frac{x^2 + x + 1}{-x^2 + x + 2} = \frac{(-3)^2 + (-3) + 1}{-(-3)^2 + (-3) + 2} = \frac{9 - 3 + 1}{-9 - 3 + 2} = \frac{7}{-10} = -\frac{7}{10}. \]

When direct substitution leads to 0/0, it’s a signal that the function might have a “hole” at that point, and we can often simplify the expression to reveal the true limit.

The most common technique for this is factoring and canceling. If plugging in ‘a’ makes both numerator and denominator zero, then (x-a) must be a factor of both.

In Example 4.9.2, factoring makes the common (x+2) factor explicit. Since we’re looking at a limit as x approaches -2, not at -2, we can cancel (x+2) as long as x is not exactly -2. After cancellation, the new function is identical to the original everywhere except at x=-2, but it is now continuous at x=-2, allowing direct substitution.

Example 4.9.3 demonstrates the same principle with higher-degree polynomials. If you have trouble factoring, remember the Factor Theorem: if P(a)=0, then (x-a) is a factor of P(x). You can then use polynomial long division or synthetic division to find the other factors.

4.9.2 Rationalization

We can use the difference of squares factorization, \(a^2 - b^2 = (a - b)(a + b)\), especially for expressions involving square roots.

Example 4.9.4 / 4.9.5:

Find \(\lim_{x \to 1^+}\frac{\sqrt{2 - x} - 1}{x - 1}\).

• Question: Can you evaluate the limit by substitution? Substituting \(x=1\) gives \(\frac{\sqrt{2-1} - 1}{1-1} = \frac{\sqrt{1} - 1}{0} = \frac{0}{0}\), an indeterminate form. So, no direct substitution.

4.9.2 Rationalization

• Solution: Multiply by the conjugate of the numerator: \[ \begin{aligned} \lim_{x \to 1^+} \frac{\sqrt{2 - x} - 1}{x - 1} &= \lim_{x \to 1^+} \frac{\sqrt{2 - x} - 1}{x - 1} \cdot \frac{\sqrt{2 - x} + 1}{\sqrt{2 - x} + 1} & \text{(Multiply by conjugate)} \\ &= \lim_{x \to 1^+} \frac{(\sqrt{2 - x})^2 - (1)^2}{(x - 1)(\sqrt{2 - x} + 1)} & \text{(Difference of squares)} \\ &= \lim_{x \to 1^+} \frac{(2 - x) - 1}{(x - 1)(\sqrt{2 - x} + 1)} \\ &= \lim_{x \to 1^+} \frac{1 - x}{(x - 1)(\sqrt{2 - x} + 1)} \\ &= \lim_{x \to 1^+} \frac{-(x - 1)}{(x - 1)(\sqrt{2 - x} + 1)} & \text{(Factor out -1 from numerator)} \\ &= \lim_{x \to 1^+} \frac{-1}{\sqrt{2 - x} + 1} & \text{(Cancel common factor (x-1))} \\ &= \frac{-1}{\sqrt{2 - 1} + 1} & \text{(Now substitute, denominator is not zero)} \\ &= \frac{-1}{1 + 1} = -\frac{1}{2}. \end{aligned} \]

Rationalization is another key algebraic technique for evaluating limits, especially when dealing with expressions involving square roots that lead to the 0/0 indeterminate form.

The trick is to multiply both the numerator and the denominator by the conjugate of the term containing the square root. The conjugate simply flips the sign between the two terms (e.g., the conjugate of (A - B) is (A + B)).

When you multiply an expression by its conjugate, you use the difference of squares formula, (A-B)(A+B) = A^2 - B^2. This eliminates the square root from that part of the expression, often leading to a simplification that allows cancellation of the problematic (x-a) term.

In the example, after rationalizing, we found a common factor (x-1) (or (1-x)) that could be canceled, leading to a function that is continuous at x=1, allowing direct substitution. Remember to be careful with signs when factoring out -1!

4.9.3 Squeeze Theorem

The Squeeze Theorem is a powerful tool for finding the limit of a function that is “squeezed” between two other functions whose limits are known and equal.

Important

Theorem 4.9.6 (Squeeze Theorem)

Suppose \(f, g\) and \(h\) are defined on an open interval \(I\) containing \(a\), except possibly at \(x = a\).

If \(f(x) \leq g(x) \leq h(x)\) on \(I\) (except possibly at \(x = a\)), and \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\), then \[ \lim_{x \to a} g(x) = L. \]

(Also known as the Sandwich Theorem or Pinching Theorem)

4.9.3 Squeeze Theorem

Example 4.9.7: Evaluate \(\lim_{x \to 0} x^2 \sin \left(\frac{1}{x^2}\right)\).

Common Wrong Solution:

\(\lim_{x \to 0} x^2 \sin \left(\frac{1}{x^2}\right) \stackrel{*}{=} \lim_{x \to 0} x^2 \cdot \lim_{x \to 0} \sin \left(\frac{1}{x^2}\right) = 0 \cdot (\text{DNE}) \neq 0\).

The product rule for limits is wrongly applied because \(\lim_{x \to 0} \sin \left(\frac{1}{x^2}\right)\) does not exist (it oscillates).

4.9.3 Squeeze Theorem

Correct Solution (using Squeeze Theorem):

1. Find bounds for the oscillating part: We know that for any value \(k\), \(-1 \leq \sin(k) \leq 1\).

   So, for \(x \neq 0\), \(-1 \leq \sin \left(\frac{1}{x^2}\right) \leq 1\).

2. Multiply by the non-negative factor: Since \(x^2 \geq 0\) for all \(x\), we can multiply the inequality by \(x^2\) without flipping the signs: \[ \underbrace{-x^2}_{f(x)} \leq \underbrace{x^2 \sin \left(\frac{1}{x^2}\right)}_{g(x)} \leq \underbrace{x^2}_{h(x)}. \]

3. Evaluate limits of the bounding functions:

   \(\lim_{x \to 0} (-x^2) = 0\)

   \(\lim_{x \to 0} (x^2) = 0\)

4. Apply the Squeeze Theorem: Since \(g(x)\) is squeezed between \(f(x)\) and \(h(x)\), and both \(f(x)\) and \(h(x)\) approach \(0\) as \(x \to 0\), we conclude: \[ \lim_{x \to 0} x^2 \sin \left(\frac{1}{x^2}\right) = 0. \]

The Squeeze Theorem is a brilliant technique when you’re dealing with a function that oscillates or whose limit is hard to determine directly, but you can “trap” it between two simpler functions.

The core idea is to find two other functions, \(f(x)\) and \(h(x)\), such that \(f(x)\) is always less than or equal to your function \(g(x)\), and \(h(x)\) is always greater than or equal to \(g(x)\). If both \(f(x)\) and \(h(x)\) approach the same limit \(L\) as \(x\) approaches \(a\), then \(g(x)\) must also approach \(L\). It’s like pinching a piece of bread between two fingers; if the fingers meet at a point, the bread must be at that point too.

The example of \(x^2 \sin(1/x^2)\) is a classic illustration. The \(\sin(1/x^2)\) part oscillates wildly near zero, so its limit does not exist. This means we cannot use the product rule for limits directly. However, we know that sine functions are always bounded between -1 and 1. By multiplying this inequality by \(x^2\) (which is non-negative), we create two new bounding functions, \(-x^2\) and \(x^2\). Both of these functions clearly go to 0 as \(x\) goes to 0. Therefore, the function in the middle must also go to 0.

4.10 One-sided Limits in Practice

Recall the relationship between a two-sided limit and one-sided limits:

Important

Theorem 4.10.1 (Equal One-sided Limits)

\(\lim_{x \to c} f(x) = L\) if and only if \(\lim_{x \to c^-} f(x) = L\) and \(\lim_{x \to c^+} f(x) = L\).

This theorem is especially useful for piecewise functions.

4.10 One-sided Limits in Practice

Example 4.10.2: Consider the function \[ f(x) = \left\{ \begin{array}{ll} -x & \text{if } x < -1 \\ x^2 & \text{if } |x| \leq 1, \text{ i.e., } -1 \leq x \leq 1 \\ 2 & \text{if } x > 1 \end{array} \right. \]

Determine if the following limits exist and their values:

1. \(\lim_{x \to -1^-} f(x)\)
2. \(\lim_{x \to -1^+} f(x)\)
3. \(\lim_{x \to -1} f(x)\)
4. \(\lim_{x \to 1} f(x)\)

Graph of piecewise function

4.10 One-sided Limits in Practice

Solution:

1. \(\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-x) = -(-1) = 1\).
2. \(\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (x^2) = (-1)^2 = 1\).
3. Since \(\lim_{x \to -1^-} f(x) = 1\) and \(\lim_{x \to -1^+} f(x) = 1\), the limit \(\lim_{x \to -1} f(x)\) exists and equals \(1\).
4. For \(\lim_{x \to 1} f(x)\):
   • \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2) = (1)^2 = 1\).
   • \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2) = 2\). Since \(\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)\), the limit \(\lim_{x \to 1} f(x)\) does not exist.

This example perfectly demonstrates the practical use of the Equal One-sided Limits Theorem for analyzing piecewise functions.

We need to be careful to choose the correct function definition based on whether we are approaching from the left or the right.

At x = -1:

• Approaching from the left (x < -1), we use the ‘-x’ definition. The limit is 1.
• Approaching from the right (x >= -1), we use the ‘x^2’ definition. The limit is 1.

Since both one-sided limits are equal, the overall limit at x=-1 exists and is 1.

At x = 1:

• Approaching from the left (x <= 1), we use the ‘x^2’ definition. The limit is 1.
• Approaching from the right (x > 1), we use the ‘2’ definition. The limit is 2.

Since the left and right-hand limits are not equal, the overall limit at x=1 does not exist. This indicates a jump discontinuity at x=1.

This systematic approach is essential for handling piecewise functions and understanding their behavior at the transition points.

4.11 Limit Laws for Infinite Limits

When dealing with infinite limits, some standard limit laws need to be adapted or interpreted carefully.

Important

Theorem 4.11.1 (Limit Laws for Infinite Limits)

Suppose that \(\lim_{x \to a} f(x) = \infty\), \(\lim_{x \to a} g(x) = \infty\), and \(\lim_{x \to a} h(x) = c\) (where \(c\) is a constant). Then:

1. \(\lim_{x \to a}(f(x) + g(x)) = \infty\); \(\lim_{x \to a}(f(x) + h(x)) = \infty\). (Infinity plus infinity is infinity; infinity plus a finite number is infinity)

2. \(\lim_{x \to a}f(x) \cdot g(x) = \infty\). (Infinity times infinity is infinity)

3. \(\lim_{x \to a}f(x) \cdot h(x) = \infty\) if \(c > 0\) and \(\lim_{x \to a}f(x) \cdot h(x) = -\infty\) if \(c < 0\).

   (Infinity times a positive constant is infinity; infinity times a negative constant is negative infinity)

4. \(\lim_{x \to a}\frac{1}{f(x)} = 0\).

   (One divided by an infinitely large number approaches zero)

The same laws hold for one-sided limits (\(\lim_{x \to a^+}\) and \(\lim_{x \to a^-}\)).

4.11 Limit Laws for Infinite Limits

Example 4.11.2:

Evaluate \(\lim_{x \to \pi /2^-}(\tan x + 2\sin x)\).

• \(\lim_{x \to \pi /2^-}(\tan x) = +\infty\).
• \(\lim_{x \to \pi /2^-}(2\sin x) = 2\sin (\pi /2) = 2 \cdot 1 = 2\).
• By Law (1): \(\lim_{x \to \pi /2^-}(\tan x + 2\sin x) = +\infty + 2 = +\infty\).

Example 4.11.3: Evaluate \(\lim_{x \to \pi /2^-}(-3\tan x\sin x)\).

• \(\lim_{x \to \pi /2^-}(\tan x) = +\infty\).
• \(\lim_{x \to \pi /2^-}(-3\sin x) = -3\sin (\pi /2) = -3 \cdot 1 = -3\).
• By Law (3) (with \(c < 0\)): \(\lim_{x \to \pi /2^-}(-3\tan x\sin x) = -\infty\).

Example 4.11.4: Evaluate \(\lim_{x \to \pi /2^+}\frac{x}{\tan x}\).

• \(\lim_{x \to \pi /2^+}x = \pi /2\).
• \(\lim_{x \to \pi /2^+}\tan x = -\infty\).
• This can be rewritten as \(\lim_{x \to \pi /2^+} x \cdot \frac{1}{\tan x}\).
• By Law (4), \(\lim_{x \to \pi /2^+}\frac{1}{\tan x} = 0\).
• By product law (finite limit times zero): \(\lim_{x \to \pi /2^+} \frac{x}{\tan x} = \frac{\pi}{2} \cdot 0 = 0\).

These laws help us manipulate and combine infinite limits, providing a way to determine the overall behavior of expressions involving functions that tend towards infinity.

It’s crucial to understand the interpretations: for instance, adding a finite number to infinity still results in infinity. Multiplying infinity by a positive constant yields infinity, while multiplying by a negative constant results in negative infinity. And one divided by infinity is zero.

The examples illustrate these rules:

• For (tan x + 2 sin x), tan x goes to positive infinity, and 2 sin x goes to a finite value. Their sum is positive infinity.
• For (-3 tan x sin x), tan x goes to positive infinity, but -3 sin x goes to a negative finite value (-3). Their product is negative infinity.
• For x / tan x, the numerator approaches a finite value, and the denominator approaches negative infinity. This is equivalent to multiplying the numerator by 1/denominator, where 1/denominator approaches 0. So the overall limit is 0.

These rules are essential for analyzing functions near vertical asymptotes and understanding how different parts of an expression contribute to its infinite behavior.

Some Useful Techniques for Infinite Limits

When we have \(\lim_{x \to a} f(x) = 0\), the limit of \(\frac{1}{f(x)}\) will diverge (either to \(\infty\), \(-\infty\), or not exist).

How do we know which one will hold?

Important

Theorem 4.11.5: Suppose \(\lim_{x \to a} f(x) = 0\).

1. If \(f(x) > 0\) on some deleted neighborhood of \(a\), then \(\lim_{x \to a} \frac{1}{f(x)} = \infty\).
2. If \(f(x) < 0\) on some deleted neighborhood of \(a\), then \(\lim_{x \to a} \frac{1}{f(x)} = -\infty\).
3. Otherwise, \(\lim_{x \to a} \frac{1}{f(x)}\) does not exist (e.g., if \(f(x)\) changes sign around \(a\)).

(A deleted neighborhood of \(a\) is \((a-\delta, a) \cup (a, a+\delta)\).)

Some Useful Techniques for Infinite Limits

One-sided Limits Version

For one-sided limits, we have similar results:

Important

Proposition 4.11.6:

Suppose \(\lim_{x \to a^+} f(x) = 0\).

1. If \(f(x) > 0\) for \(x \in (a, a + \delta)\), then \(\lim_{x \to a^+} \frac{1}{f(x)} = \infty\).
2. If \(f(x) < 0\) for \(x \in (a, a + \delta)\), then \(\lim_{x \to a^+} \frac{1}{f(x)} = -\infty\).

(Similar rules apply for \(\lim_{x \to a^-} f(x) = 0\) by considering \(x \in (a - \delta, a)\).)

Example 4.11.7:

Evaluate \(\lim_{x \to 1^+}\frac{1}{1 - x^3}\).

• Note that \(\lim_{x \to 1^+} (1 - x^3) = 0\).
• For \(x > 1\) (approaching from the right), \(x^3 > 1\), so \(1 - x^3 < 0\).
• Therefore, by Proposition 4.11.6 (b), \(\lim_{x \to 1^+}\frac{1}{1 - x^3} = -\infty\).

Some Useful Techniques for Infinite Limits

One-sided Limits Version

Example 4.11.8:

Evaluate \(\lim_{x \to -2^-}\frac{x - 1}{x + 2}\).

• Note that \(\lim_{x \to -2^-} (x - 1) = -2 - 1 = -3\).
• Note that \(\lim_{x \to -2^-} (x + 2) = 0\).
• For \(x < -2\) (approaching from the left), \(x + 2 < 0\).
• So, \(\lim_{x \to -2^-}\frac{1}{x + 2} = -\infty\).
• Therefore, \(\lim_{x \to -2^-}\frac{x - 1}{x + 2} = (-3) \cdot (-\infty) = \infty\).

These theorems provide a systematic way to determine the sign of an infinite limit when the denominator approaches zero. This is crucial when evaluating limits that result in a finite number divided by zero.

The key is to analyze the sign of the denominator (or the function \(f(x)\)) as \(x\) approaches \(a\).

• If \(f(x)\) is positive when approaching ‘a’, then 1/f(x) goes to positive infinity.
• If \(f(x)\) is negative when approaching ‘a’, then 1/f(x) goes to negative infinity.
• If \(f(x)\) changes sign (e.g., \(1/x\) as \(x \to 0\)), then the two-sided limit does not exist.

The one-sided versions are particularly helpful for functions with vertical asymptotes where the function approaches different infinities from each side.

In Example 4.11.7, as \(x \to 1^+\) means \(x\) is slightly greater than 1, so \(x^3\) is slightly greater than 1, making \(1-x^3\) a small negative number. Hence the limit is negative infinity.

In Example 4.11.8, the numerator goes to -3. The denominator (x+2) goes to 0 as \(x \to -2^-\). Since \(x\) is slightly less than -2, (x+2) is a small negative number. So 1/(x+2) goes to negative infinity. Then a negative number (-3) multiplied by negative infinity gives positive infinity.

4.12 Evaluation of Limits at Infinity

Limit theorems and laws also hold for \(\lim_{x \to \infty}\) and \(\lim_{x \to -\infty}\), provided the respective limits exist.

For example, if \(\lim_{x \to \infty} f(x) = L\) and \(\lim_{x \to \infty} g(x) = M\) exist, then: \[ \lim_{x \to \infty} (f(x) + g(x)) = L + M \] \[ \lim_{x \to \infty} (f(x) \cdot g(x)) = L \cdot M \] \[ \lim_{x \to \infty} (f(x))^n = L^n, \text{ for } n \in \mathbb{Z}^+. \]

4.12 Evaluation of Limits at Infinity

Some Useful Limits

Important

Theorem 4.12.1:

1. If \(n\) is a positive integer, then \(\lim_{x \to \infty} \frac{1}{x^n} = 0\) and \(\lim_{x \to -\infty} \frac{1}{x^n} = 0\).
2. If \(m\) and \(n\) are positive integers, then \(\lim_{x \to \infty} \frac{1}{x^{m/n}} = 0\) and \(\lim_{x \to -\infty} \frac{1}{x^{m/n}} = 0\), provided \(n\) is an odd integer for \(x \to -\infty\).

Example 4.12.2:

1. \(\lim_{x \to \infty}\frac{1}{x^5} = 0.\)
2. \(\lim_{x \to -\infty}\frac{1}{x^{7 / 3}} + e^x = 0.\) (Note: \(\lim_{x \to -\infty} e^x = 0\))
3. \(\lim_{x \to \infty}\frac{1}{x^{3 / 4}} = 0.\)

Evaluating limits at infinity involves understanding the behavior of functions as ‘x’ gets extremely large, positively or negatively. The fundamental limit laws we discussed earlier still apply here, making these calculations systematic.

Theorem 4.12.1 is particularly important. It states that any power of ‘1/x’ (where the exponent is positive) will approach zero as ‘x’ goes to positive or negative infinity. This is because the denominator grows infinitely large, making the fraction infinitesimally small.

Part (b) extends this to fractional exponents (roots), with a caveat for negative infinity: for even roots (like square roots), the base must be non-negative. So, for \(\lim_{x \to -\infty} 1/x^{m/n}\) to be defined and 0, ‘n’ must be an odd integer to allow negative x values.

These basic results are the building blocks for evaluating limits of rational functions and other more complex expressions at infinity.

Evaluating Limits at Infinity for Rational Functions

A common technique for rational functions at infinity is to divide both the numerator and denominator by the highest power of \(x\) in the denominator.

Example 4.12.3:

Evaluate \(\lim_{x \to \infty}\frac{x + 4}{x^2 - 6x + 5}\).

Solution:

Divide numerator and denominator by \(x^2\) (highest power of \(x\) in denominator): \[ \begin{aligned} \lim_{x \to \infty} \frac{x + 4}{x^2 - 6x + 5} &= \lim_{x \to \infty} \frac{\frac{x}{x^2} + \frac{4}{x^2}}{\frac{x^2}{x^2} - \frac{6x}{x^2} + \frac{5}{x^2}} \\ &= \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{4}{x^2}}{1 - \frac{6}{x} + \frac{5}{x^2}} \\ &= \frac{\lim_{x \to \infty} \frac{1}{x} + \lim_{x \to \infty} \frac{4}{x^2}}{\lim_{x \to \infty} 1 - \lim_{x \to \infty} \frac{6}{x} + \lim_{x \to \infty} \frac{5}{x^2}} & \text{(Apply limit laws)} \\ &= \frac{0 + 0}{1 - 0 + 0} = \frac{0}{1} = 0. \end{aligned} \]

Evaluating Limits at Infinity for Rational Functions

Example 4.12.4:

Evaluate \(\lim_{x \to \infty}\frac{x^3 + 4x - 5}{7x^3 - 6x + 5}\).

Solution:

Divide numerator and denominator by \(x^3\) (highest power of \(x\) in denominator): \[ \begin{aligned} \lim_{x \to \infty} \frac{x^3 + 4x - 5}{7x^3 - 6x + 5} &= \lim_{x \to \infty} \frac{\frac{x^3}{x^3} + \frac{4x}{x^3} - \frac{5}{x^3}}{\frac{7x^3}{x^3} - \frac{6x}{x^3} + \frac{5}{x^3}} \\ &= \lim_{x \to \infty} \frac{1 + \frac{4}{x^2} - \frac{5}{x^3}}{7 - \frac{6}{x^2} + \frac{5}{x^3}} \\ &= \frac{1 + 0 - 0}{7 - 0 + 0} = \frac{1}{7}. \end{aligned} \]

The technique of dividing by the highest power of ‘x’ in the denominator is the standard and most reliable way to evaluate limits of rational functions at infinity.

This technique works because it transforms the expression into a sum of terms where ‘x’ is in the denominator. As ‘x’ approaches infinity, all these terms (like 1/x, 4/x^2, etc.) approach zero, leaving only the constant terms or the coefficients of the highest powers.

In Example 4.12.3, the highest power in the denominator is \(x^2\). After dividing, all terms in the numerator and denominator except for the constant ‘1’ in the denominator go to zero, resulting in a limit of 0. This means the x-axis (y=0) is a horizontal asymptote.

In Example 4.12.4, the highest power in the denominator is \(x^3\). After dividing, the only remaining non-zero terms are the coefficients of \(x^3\) in the numerator and denominator, which are 1 and 7, respectively. The limit is 1/7. This means y=1/7 is a horizontal asymptote.

This method effectively compares the growth rates of the numerator and denominator polynomials.

Limits of Hyperbolic Functions

Hyperbolic functions also have specific behaviors at infinity.

Important

Proposition 4.12.5:

1. \(\lim_{x \to \infty} \sinh x = +\infty\) and \(\lim_{x \to -\infty} \sinh x = -\infty\).
2. \(\lim_{x \to \infty} \cosh x = +\infty\) and \(\lim_{x \to -\infty} \cosh x = +\infty\).
3. \(\lim_{x \to \infty} \tanh x = +1\) and \(\lim_{x \to -\infty} \tanh x = -1\).

Hyperbolic functions are defined in terms of exponential functions, so their limits at infinity are derived from the limits of \(e^x\) and \(e^{-x}\).

For \(\sinh x\): As x goes to positive infinity, \(e^x\) dominates, so it goes to positive infinity. As x goes to negative infinity, \(-e^{-x}\) dominates, so it goes to negative infinity.

For \(\cosh x\): As x goes to either positive or negative infinity, both \(e^x\) and \(e^{-x}\) (or one of them) contribute positively, so it always goes to positive infinity.

For \(\tanh x\): This function has horizontal asymptotes. As x goes to positive infinity, we divide by \(e^x\) and see it approaches 1. As x goes to negative infinity, we divide by \(e^{-x}\) and see it approaches -1. These proofs illustrate the algebraic manipulation necessary.

Understanding these limits is important for applications in physics and engineering, especially when dealing with phenomena that involve exponential growth or decay.

Limits Involving Square Roots

When evaluating limits at infinity involving square roots, be careful with the definition of \(\sqrt{x^2}\).

Example 4.12.6: Evaluate \(\lim_{x \to \infty}\frac{\sqrt{2x^2 + 1}}{3x - 5}\).

Solution: For \(x > 0\), we know \(x = \sqrt{x^2}\). Divide numerator and denominator by \(x\): \[ \begin{aligned} \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} &= \lim_{x \to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{3x - 5}{x}} \\ &= \lim_{x \to \infty} \frac{\sqrt{\frac{2x^2 + 1}{x^2}}}{3 - \frac{5}{x}} & \text{(Since } x > 0, x = \sqrt{x^2}) \\ &= \lim_{x \to \infty} \frac{\sqrt{2 + \frac{1}{x^2}}}{3 - \frac{5}{x}} \\ &= \frac{\sqrt{2 + 0}}{3 - 0} = \frac{\sqrt{2}}{3}. \end{aligned} \]

Limits Involving Square Roots

Example 4.12.7: Evaluate \(\lim_{x \to -\infty}\frac{\sqrt{2x^2 + 1}}{3x - 5}\).

Solution: For \(x < 0\), we know \(x = -\sqrt{x^2}\). Divide numerator and denominator by \(x\): \[ \begin{aligned} \lim_{x \to -\infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} &= \lim_{x \to -\infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{3x - 5}{x}} \\ &= \lim_{x \to -\infty} \frac{\frac{\sqrt{2x^2 + 1}}{-\sqrt{x^2}}}{3 - \frac{5}{x}} & \text{(Since } x < 0, x = -\sqrt{x^2}) \\ &= \lim_{x \to -\infty} \frac{-\sqrt{\frac{2x^2 + 1}{x^2}}}{3 - \frac{5}{x}} \\ &= \lim_{x \to -\infty} \frac{-\sqrt{2 + \frac{1}{x^2}}}{3 - \frac{5}{x}} \\ &= \frac{-\sqrt{2 + 0}}{3 - 0} = -\frac{\sqrt{2}}{3}. \end{aligned} \]

Limits at infinity involving square roots require careful handling of the term ‘x’ when moving it inside or outside a square root.

The critical point is that \(\sqrt{x^2}\) is equal to \(|x|\).

• If \(x > 0\) (as in \(\lim_{x \to \infty}\)), then \(|x| = x\), so we can use \(x = \sqrt{x^2}\).
• If \(x < 0\) (as in \(\lim_{x \to -\infty}\)), then \(|x| = -x\), which means \(x = -\sqrt{x^2}\). This sign difference is crucial.

In Example 4.12.6, for positive infinity, we divide by x and move it into the square root as \(x^2\). This yields a positive limit.

In Example 4.12.7, for negative infinity, when we divide by x in the numerator, we must write \(x = -\sqrt{x^2}\). This introduces a negative sign outside the square root, leading to a negative limit.

Failing to account for the sign of ‘x’ when working with square roots at negative infinity is a common error, so pay close attention!

4.13 Continuous Functions

We often focus on functions that are continuous over entire intervals.

Intervals and End-points

Consider intervals like \((a, b)\), \([a, b)\), \((a, b]\), \([a, b]\), \((a, \infty)\), \([a, \infty)\), \((-\infty, b)\), \((-\infty, b]\).

• Points \(a\) and \(b\) are endpoints. \(a\) is a left endpoint, \(b\) is a right endpoint.
• A point \(x\) in an interval is an interior point if it is not an endpoint.

Continuity on an Interval

Important

Definition 4.13.1: The function \(f\) is said to be continuous on the interval \(I\) if:

1. \(f\) is continuous at every interior point \(c\) of \(I\), i.e., \(\lim_{x \to c} f(x) = f(c)\).
2. If the left endpoint \(a\) of \(I\) is included in \(I\), \(f\) is continuous from the right there, i.e., \(\lim_{x \to a^+} f(x) = f(a)\).
3. If the right endpoint \(b\) of \(I\) is included in \(I\), \(f\) is continuous from the left there, i.e., \(\lim_{x \to b^-} f(x) = f(b)\).

The graph of a continuous function on an interval will be a curve with no breaks and no jumps.

4.13 Continuous Functions

Example 4.13.2:

The Heaviside function \(H(x) = \left\{ \begin{array}{ll} 0 & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{array} \right.\)

• Is continuous on the interval \([0, \infty)\) (continuous from the right at \(x=0\)).
• Is NOT continuous on \((-\infty, \infty)\) (due to jump at \(x=0\)).
• Is NOT continuous on \((-\infty, 0]\) (not continuous from the left at \(x=0\)).

Continuity on an interval is a stronger condition than continuity at a single point. It means the function behaves “nicely” throughout the entire interval.

The definition for continuity on an interval has three parts:

1. For points within the interval (interior points), the standard two-sided continuity condition must hold.
2. If the interval includes its left endpoint, the function must be continuous from the right at that endpoint. This makes sense because the function is only defined to the right of it.
3. Similarly, if the interval includes its right endpoint, the function must be continuous from the left at that endpoint.

Graphically, this means you can trace the entire function over that interval without lifting your pencil.

The Heaviside function example helps illustrate this. While it has a jump at x=0, it is continuous on \([0, \infty)\) because it is continuous from the right at 0 and continuous at all interior points (which are constant). However, it’s not continuous on \((-\infty, \infty)\) or \((-\infty, 0]\) due to the jump.

Some Known Continuous Functions

Many familiar functions are continuous on their domains:

• Polynomials: E.g., \(f(x) = x^3 + x + 1\) (continuous on \(\mathbb{R}\)).
• \(n^{th}\)-root functions: E.g., \(g(x) = \sqrt[5]{x}\) (continuous on \(\mathbb{R}\)).
• Trigonometric functions: E.g., \(\sin x, \cos x\) (continuous on \(\mathbb{R}\)).
• Exponential and Logarithmic functions: E.g., \(e^x, 3^x\) (continuous on \(\mathbb{R}\)); \(\ln x\) (continuous on \((0, \infty)\)).
• Hyperbolic functions: E.g., \(\sinh x, \cosh x\) (continuous on \(\mathbb{R}\)).

For these functions, to determine \(\lim_{x \to c} f(x)\) at a point \(x=c\) in their domain, we can simply substitute \(c\) into the function.

Some Known Continuous Functions

Combination and Composite Functions

Important

Theorem 4.13.3 (Combination of Continuous Functions)

If functions \(f\) and \(g\) are continuous on a set \(S\), then \(f \pm g\), \(f \cdot g\), and \(f/g\) (provided \(g(c) \neq 0\)) are also continuous on \(S\).

Proposition 4.13.5 (Composite of Continuous Functions)

If \(f\) is continuous on its domain \(D_f\) and \(g\) is continuous on the range \(R_f\) of \(f\), then the composite function \(g \circ f\) is continuous on \(D_f\).

Some Known Continuous Functions

Combination and Composite Functions

Example 4.13.4:

The function \(f(x) = \frac{\sin(x - e^x)}{\ln x}\) is continuous on \((1, \infty)\).

(Numerator is composite of continuous functions, denominator \(\ln x\) is continuous and non-zero on \((1, \infty)\)).

Example 4.13.6:

\(f(x) = x^3\) and \(g(x) = \sin x\) are continuous on \(\mathbb{R}\). Their composite functions \(f \circ g(x) = (\sin x)^3\) and \(g \circ f(x) = \sin(x^3)\) are also continuous on \(\mathbb{R}\).

This slide summarizes that many common functions we encounter are naturally continuous over their entire domains. This is a very useful fact because it means for any point in their domain, their limit is simply their function value.

Furthermore, continuity is “preserved” under basic arithmetic operations and composition, similar to limit laws.

• The sum, difference, product, and quotient (with non-zero denominator) of continuous functions are also continuous.
• The composition of continuous functions is also continuous. This is extremely powerful for understanding complex functions, as it means if you chain together continuous functions, the result will also be continuous (within its domain).

The examples illustrate how to apply these properties. For instance, a rational function where the numerator is a composite of continuous functions and the denominator is also a continuous function (and not zero) will be continuous. This greatly simplifies the task of determining continuity.

Composite Functions and Their Domains

Example 4.13.7:

Let \(f(x) = \sqrt{x - 4}\) on \([4, \infty)\) and \(g(x) = x^2\).

Find \(g \circ f\) and its domain where it is continuous.

Solution:

\(f\) is continuous on \([4, \infty)\). \(g(x) = x^2\) is continuous on \(\mathbb{R}\).

The composite function \(g \circ f\) is continuous on the domain where \((g \circ f)(x)\) is defined.

\((g \circ f)(x) = g(f(x)) = g(\sqrt{x - 4})\).

This requires \(x - 4 \geq 0\), so \(x \geq 4\).

Then, \(g(\sqrt{x - 4}) = (\sqrt{x - 4})^2 = x - 4\), provided \(x \geq 4\).

Thus, \((g \circ f)(x) = x - 4\) with domain \([4, \infty)\).

So, \((g \circ f)(x)\) is continuous on \([4, \infty)\).

Note

Remark 4.13.8:

Although \((g \circ f)(x)\) and \(h(x) = x - 4\) have the same expression, they are different functions because their domains differ. \((g \circ f)(x) = x - 4\) is only defined on \([4, \infty)\), while \(h(x) = x - 4\) is defined on \(\mathbb{R}\).

Composite Functions and Their Domains

Example 4.13.9:

Let \(f(x) = \sqrt{x - 4}\) on \([4, \infty)\) and \(g(x) = x^2\).

Find \(f \circ g\) and its domain where it is continuous.

Solution:

\((f \circ g)(x) = f(g(x)) = f(x^2)\).

This requires \(x^2 - 4 \geq 0\), which means \(x^2 \geq 4\).

So, \(x \leq -2\) or \(x \geq 2\).

Thus, \((f \circ g)(x) = \sqrt{x^2 - 4}\) with domain \((-\infty, -2] \cup [2, \infty)\).

The composite function \(f \circ g\) is continuous on its domain \((-\infty, -2] \cup [2, \infty)\).

Inverse of Continuous Functions

Important

Proposition 4.13.10:

If \(f\) is a one-to-one continuous function, then its inverse \(f^{-1}\) is continuous.

This implies that inverse trigonometric functions (\(\sin^{-1}x, \cos^{-1}x, \tan^{-1}x\)) are continuous on their respective restricted domains.

These examples highlight the importance of considering the domain when forming composite functions, as the domain of the composite function is often more restricted than the individual component functions. Even if the algebraic expression looks simple after simplification, the domain imposed by the original functions must be respected.

For \(g \circ f\): The inner function \(f(x) = \sqrt{x-4}\) requires \(x \geq 4\). This domain restriction carries over to the composite function. Even though \((\sqrt{x-4})^2\) simplifies to \(x-4\), the domain of \(g \circ f\) is still \([4, \infty)\).

For \(f \circ g\): The inner function \(g(x) = x^2\) has domain \(\mathbb{R}\). However, the outer function \(f\) requires its input to be \(\geq 4\). So, \(x^2 \geq 4\), which implies \(x \leq -2\) or \(x \geq 2\). This defines the domain of \(f \circ g\).

The proposition about inverse functions is also very significant. It means that if a function is continuous and one-to-one (meaning it has an inverse), then its inverse function will also be continuous. This explains why inverse trigonometric functions, when defined on appropriate restricted domains, are also continuous.

4.14 The Intermediate Value Theorem (IVT)

For a function \(f\) that is continuous on a closed and bounded interval \([a, b]\), the Intermediate Value Theorem is a powerful result.

Important

Theorem 4.14.1 (Intermediate Value Theorem)

Suppose that \(f\) is continuous on the closed and bounded interval \([a, b]\) where \(f(a) \neq f(b)\).

Let \(p\) be any real number between \(f(a)\) and \(f(b)\).

Then there exists a number \(c\) in the open interval \((a,b)\) such that \(f(c) = p\).

Remark: The number \(p\) is an intermediate value between \(f(a)\) and \(f(b)\).

4.14 The Intermediate Value Theorem (IVT)

Intermediate Value Theorem - Graphical Illustration

A typical use of the Intermediate Value Theorem is to locate a root of a function (\(f(c)=0\)).

4.14 The Intermediate Value Theorem (IVT)

IVT: Application (1) Finding Roots

If \(f\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then \(p = 0\) is an intermediate value.

By IVT, there is a real number \(c \in (a,b)\) such that \(f(c) = 0\).

Example 4.14.2:

For \(f(x) = x^3 - 5x^2 + 3\), is there a root of \(f\) in \((0,1)\)?

Solution:

1. \(f(x)\) is a polynomial, so it is continuous everywhere, including on \([0, 1]\).

2. Evaluate \(f\) at the endpoints:

   • \(f(0) = (0)^3 - 5(0)^2 + 3 = 3\).
   • \(f(1) = (1)^3 - 5(1)^2 + 3 = 1 - 5 + 3 = -1\).

3. Since \(f(0) = 3\) (positive) and \(f(1) = -1\) (negative), \(f(0)\) and \(f(1)\) have opposite signs.

4. By the Intermediate Value Theorem, there must be a value \(c \in (0, 1)\) such that \(f(c) = 0\).

   Thus, there is a root in the interval \((0, 1)\).

The Intermediate Value Theorem (IVT) is a very intuitive and powerful theorem that essentially states that a continuous function cannot “skip” any y-values between two given points. If you start at one point (a, f(a)) and end at another (b, f(b)) without lifting your pen, you must have crossed every y-value between f(a) and f(b).

The two crucial conditions for IVT are:

1. The function must be continuous on the interval.
2. The interval must be closed and bounded (\([a, b]\)).

A common application is finding roots of equations. If a continuous function has different signs at the endpoints of an interval, then it must cross the x-axis (where y=0) at least once within that interval. This is very useful for showing the existence of roots, even if we can’t easily find them algebraically.

Example 4.14.2 beautifully illustrates this. By checking the signs of f(0) and f(1), and knowing the function is continuous, we can definitively say a root exists between 0 and 1.

IVT: Application (2) Intersection of Curves

Example 4.14.3:

Use IVT to explain why the two curves \(y = \cos x\) and \(y = x^2\) intersect at some point with \(x\)-coordinate \(c\) where \(c \in (0, \frac{\pi}{2})\).

Solution:

The curves intersect if \(\cos c = c^2\), which is equivalent to \(\cos c - c^2 = 0\).

Let \(f(x) = \cos x - x^2\). We want to find a root of \(f(x)\).

1. Continuity: \(y = \cos x\) is continuous on \(\mathbb{R}\), and \(y = x^2\) is continuous on \(\mathbb{R}\).

   Therefore, \(f(x) = \cos x - x^2\) is continuous on \(\mathbb{R}\), and specifically on the closed interval \([0, \frac{\pi}{2}]\).

2. Evaluate at endpoints:

   • \(f(0) = \cos(0) - (0)^2 = 1 - 0 = 1\).
   • \(f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^2 = 0 - \frac{\pi^2}{4} = -\frac{\pi^2}{4}\).

3. Check for opposite signs: \(f(0) = 1 > 0\) and \(f\left(\frac{\pi}{2}\right) = -\frac{\pi^2}{4} \approx -2.467 < 0\).

   Since \(f(0)\) and \(f\left(\frac{\pi}{2}\right)\) have opposite signs, \(p=0\) is an intermediate value between them.

4. Apply IVT: By the Intermediate Value Theorem, there exists a real number \(c \in (0, \frac{\pi}{2})\) such that \(f(c) = 0\).

   This means \(\cos c - c^2 = 0\), or \(\cos c = c^2\), so the curves intersect at \(x=c\).

This example shows another elegant application of the IVT: proving that two functions intersect.

The strategy is to transform the intersection problem into a root-finding problem for a new function. If \(f_1(x) = f_2(x)\), then let \(f(x) = f_1(x) - f_2(x)\). If \(f(x)\) has a root, then \(f_1(x) = f_2(x)\) at that point.

In this case, we define \(f(x) = \cos x - x^2\).

1. We confirm that \(f(x)\) is continuous over the interval \([0, \pi/2]\), as it’s a difference of two continuous functions.
2. We evaluate \(f(x)\) at the endpoints. \(f(0)\) is positive, and \(f(\pi/2)\) is negative.
3. Because there’s a sign change, and the function is continuous, the IVT guarantees that \(f(x)\) must equal zero at some point \(c\) within the interval \((0, \pi/2)\).

This means the original equations \(\cos x = x^2\) must have a solution within that interval. Again, the IVT proves existence without needing to find the exact value of ‘c’.

4.15 The Extreme Value Theorem (EVT)

Another crucial result for continuous functions on closed and bounded intervals is the Extreme Value Theorem. First, let’s define extreme values.

Global Maximum/Minimum

Important

Definition 4.15.1:

1. A function \(f\) has a global maximum (absolute maximum) at \(c\) if \(f(c) \geq f(x)\) for all \(x\) in the domain \(D\). \(f(c)\) is the maximum value.
2. A function \(f\) has a global minimum (absolute minimum) at \(c\) if \(f(c) \leq f(x)\) for all \(x\) in the domain \(D\). \(f(c)\) is the minimum value.
3. The maximum and minimum values of \(f\) are called the extreme values of \(f\).

4.15 The Extreme Value Theorem (EVT)

Example 4.15.2:

Consider \(f:[-1,1]\to \mathbb{R}\) defined as \(f(x) = \left\{ \begin{array}{ll} -x & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{array} \right.\).

Does \(f\) have a global maximum and minimum on \([-1, 1]\)?

• \(f\) is not continuous at \(x=0\).
• For \(x \in [-1, 0)\), \(f(x) = -x\) means \(f(x) \in (0, 1]\). As \(x \to 0^-\), \(f(x) \to 0\).
• For \(x \in [0, 1]\), \(f(x) = 1\).
• The range of \(f\) on \([-1,1]\) is \((0, 1] \cup \{1\} = (0, 1]\).
• Global maximum is \(f(0)=1\).
• No global minimum, as \(f(x)\) approaches 0 but never reaches it in \((0,1]\).

Example 4.15.3: Let \(f(x) = \frac{1}{x}\), \(x \in (0, 1]\).

• \(f\) is continuous on \((0, 1]\) (not a closed interval).
• Global minimum is \(f(1)=1\).
• No global maximum, as \(f(x) \to \infty\) as \(x \to 0^+\).

The Extreme Value Theorem deals with the existence of highest and lowest points on a function’s graph over an interval. Before we state the theorem, we need to understand what global maximum and global minimum mean. They are the absolute highest and lowest function values that the function attains over its entire domain.

The two examples illustrate what happens when the conditions for the EVT are not met.

In Example 4.15.2, the function is not continuous at x=0. It has a global maximum at f(0)=1. However, as x approaches 0 from the left, f(x) approaches 0, but never actually reaches it. So, there is no smallest value, meaning no global minimum. This shows that continuity is important.

In Example 4.15.3, the function \(f(x) = 1/x\) is continuous on \((0, 1]\). However, the interval is not closed at 0. As x approaches 0 from the right, the function goes to infinity. So, there’s no global maximum. It does have a global minimum at x=1. This shows that the “closed interval” condition is important.

These examples demonstrate that merely having continuity or merely having a bounded interval is not enough; both conditions are required for the EVT to guarantee the existence of both a global maximum and a global minimum.

Extreme Value Theorem

The Extreme Value Theorem guarantees the existence of global maximum and minimum when both continuity and closed-and-bounded conditions are satisfied.

Important

Theorem 4.15.4 (Extreme Value Theorem)

If \(f\) is a continuous function on a closed and bounded interval \([a, b]\), then there are points \(c_1\) and \(c_2\) in \([a, b]\) such that \(f(c_1) = m\) is a global minimum and \(f(c_2) = M\) is a global maximum for \(f\).

Extreme Value Theorem - Consequence

As a consequence of both the Intermediate Value Theorem and the Extreme Value Theorem:

Tip

Corollary 4.15.5: If \(f:[a,b] \to \mathbb{R}\) is a continuous function, then the range \(R_f\) of \(f\) is a singleton \(\{f(a)\}\) or a closed and bounded interval \([f(c_1), f(c_2)]\), for some \(c_1 \in [a,b]\) and \(c_2 \in [a,b]\).

This means a continuous function on a closed interval maps that interval to another closed interval (or a single point if the function is constant).

Extreme Value Theorem

Example 4.15.6:

Determine the range of \(f(x) = 3x^2, x \in [-2,1]\).

Solution:

1. \(f(x) = 3x^2\) is a polynomial, so it is continuous on \([-2, 1]\).
2. The interval \([-2, 1]\) is closed and bounded.
3. By EVT, a global maximum and minimum exist.
4. Find critical points (where \(f'(x)=0\) or undefined) and evaluate at endpoints:
   • \(f'(x) = 6x\). Setting \(f'(x)=0\) gives \(x=0\).
   • Evaluate at endpoints and critical point:
     • \(f(-2) = 3(-2)^2 = 12\).
     • \(f(1) = 3(1)^2 = 3\).
     • \(f(0) = 3(0)^2 = 0\).
5. Global maximum is \(f(-2) = 12\). Global minimum is \(f(0) = 0\).
6. By Corollary 4.15.5, the range of \(f\) is \([0, 12]\).

The Extreme Value Theorem (EVT) is a cornerstone of calculus, guaranteeing that a continuous function on a closed interval will always attain both a highest and a lowest value within that interval. This is intuitively clear if you imagine drawing a continuous curve without lifting your pencil between two points.

The two key conditions for EVT are:

1. Continuity of the function.
2. The interval must be closed and bounded.

If these conditions are met, the EVT guarantees the existence of these extreme values, but it doesn’t tell you where they occur. Finding the exact locations (the \(c_1\) and \(c_2\)) is a task for differentiation, which we’ll cover later.

The corollary extends this idea to the range of the function. It implies that if you have a continuous function over a closed interval, its output values will also span a closed interval.

Example 4.15.6 demonstrates how to find these extreme values and thus the range of the function. We evaluate the function at the critical points within the interval and at the endpoints. The largest of these values is the global maximum, and the smallest is the global minimum.

Key Takeaways

Limits are about Approach, not always Value:

• \(\lim_{x \to a} f(x)\) describes the behavior of \(f(x)\) near \(x=a\), not necessarily at \(x=a\).
• One-sided limits help analyze behavior from left (\(\lim_{x \to a^-}\)) and right (\(\lim_{x \to a^+}\)). A two-sided limit exists if and only if one-sided limits are equal.
• Infinite limits (\(\infty, -\infty\)) describe unbounded behavior, often leading to vertical asymptotes.
• Limits at infinity describe end behavior (horizontal asymptotes).

Key Takeaways

Continuity is Key:

• A function \(f\) is continuous at \(x=c\) if \(f(c)\) is defined, \(\lim_{x \to c} f(x)\) exists, and they are equal (\(f(c) = \lim_{x \to c} f(x)\)).
• Continuous functions allow direct substitution for limits.

Key Takeaways

Powerful Theorems & Techniques:

• Limit Laws simplify evaluating limits of sums, products, quotients, and powers.
• Algebraic Techniques (factoring, rationalization) resolve \(\frac{0}{0}\) indeterminate forms.
• Squeeze Theorem finds limits of functions bounded by two other functions with the same limit.
• Intermediate Value Theorem (IVT) guarantees a function hits all values between \(f(a)\) and \(f(b)\) if continuous on \([a,b]\). Great for proving existence of roots.
• Extreme Value Theorem (EVT) guarantees continuous functions on \([a,b]\) attain global max and min.

This slide summarizes the most important concepts from the chapter.

Firstly, remember the core idea of a limit: it’s about approach, not necessarily the actual value at the point. One-sided limits are crucial for understanding this approach from different directions, and they are the building blocks for the overall limit. Infinite limits describe “blow-up” behavior (vertical asymptotes), and limits at infinity describe long-term trends (horizontal asymptotes).

Secondly, continuity ties the concept of limits to function values. A continuous function is “well-behaved”—no breaks, jumps, or holes. The immense practical benefit is that for continuous functions, finding a limit simply involves plugging in the value.

Finally, we covered powerful tools. The Limit Laws are your basic arithmetic for limits. Techniques like factoring and rationalization are your algebraic tools for tricky cases like 0/0. The Squeeze Theorem helps with oscillating functions. And the IVT and EVT are fundamental theorems that give us strong guarantees about the existence of roots and extreme values for continuous functions on closed intervals, which are incredibly useful in problem-solving and proofs. Mastering these will set you up for success in calculus.

Key Equations

Equation	Description
\(\lim_{x \to a} f(x) = L\)	The limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\).
\(\lim_{x \to a^-} f(x) = L\)	Left-hand limit: \(f(x)\) approaches \(L\) as \(x\) approaches \(a\) from values less than \(a\).
\(\lim_{x \to a^+} f(x) = L\)	Right-hand limit: \(f(x)\) approaches \(L\) as \(x\) approaches \(a\) from values greater than \(a\).
\(\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L\)	A two-sided limit exists if and only if both one-sided limits exist and are equal.

Key Equations

Equation	Description
\(\lim_{x \to a} f(x) = \infty\)	\(f(x)\) increases without bound as \(x\) approaches \(a\).
\(\lim_{x \to \infty} f(x) = L\)	\(f(x)\) approaches \(L\) as \(x\) increases without bound.
\(\lim_{x \to a} (f(x) \pm g(x)) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)\)	Limit of a sum/difference is the sum/difference of limits.
\(\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)\)	Limit of a product is the product of limits.
\(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}\), if \(\lim_{x \to a} g(x) \neq 0\)	Limit of a quotient is the quotient of limits (denominator must not be zero).
\(\lim_{x \to c} f(x) = f(c)\)	Definition of continuity at a point \(c\).

This table provides a quick reference for the most important mathematical expressions and their meanings from the chapter. These are the formulas and concepts you should commit to memory and understand deeply. They are the building blocks for all subsequent calculus topics.

Pay special attention to the conditions under which each limit law applies, especially the non-zero denominator for quotients, and the specific definitions of one-sided, infinite, and limits at infinity. The definition of continuity is also critical, as it bridges the gap between limits and function values.

Key Terms

Term	Definition
Limit of a Function	The value that a function approaches as the input (or independent variable) approaches some value.
One-sided Limit	The limit of a function as the input approaches a point from either the left (values less than the point) or the right (values greater than the point).
Left-hand Limit	The limit as \(x \to a^-\) (from the left).
Right-hand Limit	The limit as \(x \to a^+\) (from the right).
Infinite Limit	A limit in which the function’s values grow without bound (to \(\infty\)) or decrease without bound (to \(-\infty\)) as the input approaches a specific point.
Vertical Asymptote	A vertical line \(x=a\) that the graph of a function approaches as \(x\) approaches \(a\), where the function’s values tend towards \(\pm \infty\).

Key Terms

Term	Definition
Limit at Infinity	A limit in which the input (or independent variable) grows without bound (\(x \to \infty\)) or decreases without bound (\(x \to -\infty\)), and the function approaches a specific finite value \(L\).
Horizontal Asymptote	A horizontal line \(y=b\) that the graph of a function approaches as \(x \to \pm \infty\).
Indeterminate Form	An expression (like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)) whose limit cannot be determined by direct substitution, requiring further analysis or algebraic manipulation.
Continuity at a Point	A function is continuous at a point if its limit at that point exists, its value at that point exists, and these two are equal. Graphically, no breaks, jumps, or holes.
Continuity on an Interval	A function is continuous on an interval if it is continuous at every interior point, and continuous from the right/left at any included endpoints.

Key Terms

Term	Definition
Squeeze Theorem	A theorem stating that if a function \(g(x)\) is bounded between two functions \(f(x)\) and \(h(x)\) near a point \(a\), and both \(f(x)\) and \(h(x)\) have the same limit \(L\) at \(a\), then \(g(x)\) also has the limit \(L\) at \(a\).
Intermediate Value Theorem (IVT)	A theorem stating that for a continuous function on a closed interval \([a,b]\), the function takes on every value between \(f(a)\) and \(f(b)\).
Extreme Value Theorem (EVT)	A theorem stating that a continuous function on a closed and bounded interval \([a,b]\) must attain both a global maximum and a global minimum on that interval.
Global Maximum/Minimum	The absolute highest or lowest value that a function attains over its entire domain.

This glossary provides definitions for all the key terminology introduced in Chapter 4. Understanding these terms is essential for comprehending the concepts of limits and continuity.

Take time to review each term and ensure you understand its meaning in the context of calculus. Pay particular attention to the distinctions between different types of limits, the conditions for continuity, and the implications of the major theorems like IVT and EVT. These terms will be used frequently throughout your calculus studies.