Calculus I: Foundations

Limit Laws and Advanced Computations

Imron Rosyadi

Introduction to Limits: Powerful Rules for Calculus

Unlocking Complex Limit Problems

Good morning, everyone! Today, we’re diving deeper into the world of limits, specifically focusing on the powerful tools known as Limit Laws. These are not just abstract rules; they are the bedrock for understanding continuity, derivatives, and integrals, which are central to all of calculus. We’ll also explore strategies for those trickier, “indeterminate” limit problems.

Why Do We Need Limit Laws?

Breaking Down Complexity

Limit laws allow us to compute limits by breaking down complex expressions into simpler, manageable pieces.

These laws are theorems, rigorously proven from the technical definition of a limit.

Our goal: Transform difficult limit calculations into a series of easier ones.

For example, finding \(\displaystyle\lim_{x \to a} \left(3x^2 - \frac{\sin(x)}{x}\right)\) can be challenging directly.

But with limit laws, we can evaluate \(\displaystyle\lim_{x \to a} 3x^2\) and \(\displaystyle\lim_{x \to a} \frac{\sin(x)}{x}\) separately.

Why Do We Need Limit Laws?

graph LR
    A["Complex Limit: 
    lim f(x)g(x)"] --> B{Apply Limit 
    Laws};
    B --> C["Limit of 
    f(x)"];
    B --> D["Limit of 
    g(x)"];
    C --> E["Evaluate 
    lim f(x)"];
    D --> F["Evaluate 
    lim g(x)"];
    E & F --> G[Combine 
    Results];

Think of limit laws as a set of rules for algebra, but for limits. Instead of struggling with an entire complex function, we can break it apart, evaluate the limit of each piece, and then combine those results according to the laws. The Mermaid diagram visually represents this decomposition process.

The Fundamental Limit Laws

Building Blocks for Limit Evaluation

Suppose that \(\displaystyle\lim_{x \to a} f(x)\) and \(\displaystyle\lim_{x \to a} g(x)\) exist, and that \(c\) is a constant. Then:

• Sum Rule: \(\lim_{x \to a} \left(f(x)+g(x)\right) = \left(\lim_{x \to a} f(x)\right) + \left(\lim_{x \to a} g(x)\right).\)
• Difference Rule: \(\lim_{x \to a} \left(f(x)-g(x)\right) = \left(\lim_{x \to a} f(x)\right) - \left(\lim_{x \to a} g(x)\right).\)
• Constant Multiple Rule: \(\lim_{x \to a} c \cdot f(x) = c \cdot \lim_{x \to a} f(x).\)
• Product Rule: \(\lim_{x \to a} \left(f(x)\cdot g(x)\right) = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right).\)
• Quotient Rule: \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x \to a} f(x)}{\displaystyle\lim_{x \to a} g(x)}, \;\text{ if }\; \lim_{x \to a} g(x) \ne 0.\)

These rules apply to one-sided limits as well (\(x \to a^+\) or \(x \to a^-\)).

These five laws are essentially the “arithmetic” of limits. They tell us that if we know the limits of two functions, we can find the limit of their sum, difference, product, constant multiple, and quotient (as long as we’re not dividing by zero). The condition \(\lim_{x \to a} g(x) \ne 0\) for the quotient rule is crucial; it leads to an important concept we’ll discuss next.

Intuition Behind Denominator Issues

non-zero/0 vs. 0/0

What happens when the denominator approaches zero?

Case 1: \(\frac{\text{non-zero number}}{0}\)

Consider \(f(x)\) close to 4 and \(g(x)\) close to 0.

Examples: \(\frac{3.99}{0.00001} = 399000\), or \(\frac{4.01}{-0.00001} = -401000\).

This often indicates a vertical asymptote.

The limit must be \(\infty\), \(-\infty\), or DNE (if left/right limits differ).

Note

Not Indeterminate! We can determine the behavior (positive/negative infinity) by checking one-sided limits.

Intuition Behind Denominator Issues

non-zero/0 vs. 0/0

Case 2: \(\frac{0}{0}\)

When both \(f(x)\) and \(g(x)\) approach zero.

Example ratios: \(\frac{0.001}{0.0001} = 10\), or \(\frac{0.0001}{0.001} = 0.1\), or \(\frac{0.001}{-0.00001} = -100\).

This form tells us very little about the actual limit value.

Warning

Indeterminate Form! \(\frac{0}{0}\) is called an indeterminate form. We cannot determine its value without “more work.”

Understanding the difference between these two scenarios is fundamental. When the numerator is not zero but the denominator is approaching zero, the function’s value is shooting off to positive or negative infinity – we just need to determine the sign. This is predictable behavior. However, when both the numerator and the denominator are approaching zero, it’s a battle between two vanishingly small numbers, and the outcome is unclear. This is where we need specific algebraic techniques, which we’ll cover soon.

Theorem 1: Direct Substitution

When Limits Are “Easy”

If \(f\) is a polynomial or a rational function, and \(a\) is in the domain of \(f\), then \(\lim _{x\to a} f(x)=f(a).\)

This theorem is true because polynomials and rational functions are “well-behaved” (continuous) where they are defined. It’s a direct consequence of applying the limit laws repeatedly.

In practice: If you can plug \(a\) into \(f(x)\) and get a defined value (no division by zero, no square roots of negative numbers, etc.), then that value is the limit!

Example 1:

Find \(\displaystyle\lim_{x\rightarrow 3} (x^2-4)\).

Since \(f(x)=x^2-4\) is a polynomial, and \(3\) is in its domain: \(\lim_{x\rightarrow 3} (x^2-4) = (3)^2-4 = 9-4=5.\)

Theorem 1: Direct Substitution

When Limits Are “Easy”

Example 2:

Find \(\displaystyle\lim_{x\rightarrow 4}\frac{x-2}{x+2}\).

Since \(f(x)=\frac{x-2}{x+2}\) is a rational function, and \(a=4\) does not make the denominator \(0\): \(\lim_{x\rightarrow 4}\frac{x-2}{x+2}=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}.\)

Tip

Always Try Direct Substitution First! This is the simplest method. If it works, you’re done!

Theorem 1 is fantastic because it streamlines a huge number of limit problems. It means that for most simple functions you’ll encounter (polynomials, rational functions, trigonometric functions, exponentials, logarithms), if the point ‘a’ is in their domain, you can just plug it in. This highlights the concept of continuity, where the limit value equals the function’s value. The Pyodide block shows a numerical check, reinforcing that as x gets close to 3, f(x) gets close to 5.

Theorem 2: Equivalent Functions

When Functions Differ Only at a Point

If \(f(x) = g(x)\) whenever \(x \ne a\), then \(\displaystyle\lim_{x \to a} f(x) = \lim_{x \to a} g(x)\).

This theorem is crucial for many indeterminate forms (like \(\frac{0}{0}\)). It allows us to algebraically simplify a function at points other than \(a\), without changing its limit as \(x \to a\).

Theorem 2: Equivalent Functions

When Functions Differ Only at a Point

Example:

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

If we try direct substitution, we get \(\frac{1^2-1}{1-1} = \frac{0}{0}\), an indeterminate form. We need “more work.”

Notice that \(x^2-1 = (x-1)(x+1)\). So, \(\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}\).

For any \(x \ne 1\), we can cancel the \((x-1)\) terms, simplifying the expression: \(\frac{(x-1)(x+1)}{x-1} = x+1\).

Thus, \(f(x)=\frac{x^2-1}{x-1}\) is equivalent to \(g(x)=x+1\) for all \(x \ne 1\). By Theorem 2: \(\displaystyle\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} (x+1) = 1+1 = 2.\) (using Theorem 1 for \(g(x)\)).

Theorem 2 is incredibly powerful for handling indeterminate forms like 0/0. The key insight is that limits care about what happens near ‘a’, not at ‘a’. If we can simplify the expression without changing its behavior near ‘a’, then the limit remains the same. This is precisely what algebraic maneuvers like factoring and canceling do. The original function has a “hole” at x=1, while the simplified function is continuous at x=1. The limit helps us find the “height” of that hole.

Algebraic Strategy 1: Factoring

Taming \(\frac{0}{0}\) Forms with Polynomials

When direct substitution yields \(\frac{0}{0}\) for rational functions involving polynomials, factoring is your go-to strategy.

Process:

1. Attempt direct substitution. If \(\frac{0}{0}\) appears, proceed.
2. Factor the numerator and the denominator.
3. Look for common factors that cause the zero in both numerator and denominator.
4. Cancel the common factors. Remember this cancellation is valid because we are considering \(x \ne a\).
5. Evaluate the limit of the simplified expression using direct substitution.

Algebraic Strategy 1: Factoring

Taming \(\frac{0}{0}\) Forms with Polynomials

Example:

Evaluate \(\displaystyle\lim_{h \to 0} \frac{(2+h)^2 - 4}{h}\).

1. Direct substitution: \(\frac{(2+0)^2 - 4}{0} = \frac{4-4}{0} = \frac{0}{0}\). Indeterminate.
2. Expand numerator: \((2+h)^2 - 4 = (4+4h+h^2) - 4 = 4h+h^2 = h(4+h)\).
3. Rewrite the limit: \(\displaystyle\lim_{h \to 0} \frac{h(4+h)}{h}\).
4. Cancel common factor \(h\) (since \(h \ne 0\) for the limit): \(\displaystyle\lim_{h \to 0} (4+h)\).
5. Direct substitute \(h=0\): \(4+0 = 4\).

Thus, \(\displaystyle\lim_{h \to 0} \frac{(2+h)^2 - 4}{h} = 4\).

Important

“Do More Work” through Factoring! Factoring often reveals the underlying behavior of the function, allowing us to resolve the indeterminate form.

This example for factoring is canonical in calculus, as it’s the very definition of the derivative for \(f(x)=x^2\) at \(x=2\). It perfectly illustrates how factoring removes the “problematic” term that causes the 0/0 form, allowing the true limit to emerge.

Algebraic Strategy 2: Fractions

Simplifying Indeterminate Forms Involving Rational Expressions

You might encounter limits of the form \(\infty-\infty\) or \(\frac{0}{0}\) when dealing with sums or differences of fractions.

The key is to combine these fractions into a single rational expression.

Process:

1. Identify the indeterminate form.
2. Find a common denominator for the fractions.
3. Combine the fractions into a single rational expression.
4. Simplify the numerator and denominator. This often leads to a factor that can be canceled.
5. Evaluate the limit of the simplified expression.

Algebraic Strategy 2: Fractions

Simplifying Indeterminate Forms Involving Rational Expressions

Example 1: \(\infty-\infty\) form

Evaluate \(\displaystyle\lim_{t\to{0}}\left(\frac{1}{t} - \frac{1}{t^2+t}\right)\).

1. As \(t \to 0^+\), \(\frac{1}{t} \to \infty\) and \(\frac{1}{t^2+t} = \frac{1}{t(t+1)} \to \infty\). So, \(\infty-\infty\).
2. Common denominator is \(t(t+1)\). \(\frac{1}{t} - \frac{1}{t(t+1)} = \frac{t+1}{t(t+1)} - \frac{1}{t(t+1)} = \frac{(t+1)-1}{t(t+1)}.\)
3. Simplify: \(\frac{t}{t(t+1)}\).
4. Cancel \(t\) (since \(t \ne 0\)): \(\frac{1}{t+1}\).
5. \(\displaystyle\lim_{t\to{0}}\frac{1}{t+1} = \frac{1}{0+1}=1.\)

Algebraic Strategy 2: Fractions

Simplifying Indeterminate Forms Involving Rational Expressions

Example 2: \(\frac{0}{0}\) form

Evaluate \(\displaystyle\lim_{x\to{-4}}\frac{\frac{1}{4} + \frac{1}{x}}{4+x}\).

1. As \(x \to -4\), numerator \(\frac{1}{4} + \frac{1}{-4} = 0\), denominator \(4+(-4)=0\). So, \(\frac{0}{0}\).
2. Common denominator for numerator is \(4x\). \(\frac{1}{4} + \frac{1}{x} = \frac{x}{4x} + \frac{4}{4x} = \frac{x+4}{4x}.\)
3. Rewrite: \(\frac{\frac{x+4}{4x}}{4+x} = \frac{x+4}{4x(4+x)}.\)
4. Cancel \((x+4)\) (since \(x \ne -4\)): \(\frac{1}{4x}\).
5. \(\displaystyle\lim_{x\to{-4}}\frac{1}{4x} = \frac{1}{4(-4)}=-\frac{1}{16}.\)

Combining fractions is a classic algebraic manipulation trick. The goal is always the same: get rid of the factor that’s causing the indeterminate 0/0 or infinity - infinity problem. Once you’ve combined the fractions, you’ll often find a common factor you can cancel, leading back to a form where direct substitution works. This strategy is another example of “doing more work” to resolve the limit.

Limits with Absolute Values

Piecewise Definitions and Case Analysis

Recall the definition of the absolute value: \[ |a|=\begin{cases}a &\text { if } a\ge 0;\\-a&amp;\text { if } a< 0.\end{cases} \]

This definition extends to functions: \[ |f(x)|=\begin{cases} f(x), &\text{ if } f(x)\ge0;\\ -f(x), &amp;\text{ if } f(x)<0.\end{cases} \]

Working with absolute values almost always involves considering different cases based on where the expression inside the absolute value is positive or negative.

Limits with Absolute Values

Piecewise Definitions and Case Analysis

Example:

Evaluate \(\displaystyle\lim_{x \to 0} \frac{|x|}{x}\).

• If \(x > 0\), then \(|x|=x\), so \(\frac{|x|}{x} = \frac{x}{x}=1\). Thus, \(\lim_{x \to 0^+} \frac{|x|}{x} = 1\).
• If \(x < 0\), then \(|x|=-x\), so \(\frac{|x|}{x} = \frac{-x}{x}=-1\). Thus, \(\lim_{x \to 0^-} \frac{|x|}{x} = -1\).

Since the left-hand limit (\(\neq -1\)) and right-hand limit (\(\neq 1\)) are not equal, \(\displaystyle\lim_{x \to 0} \frac{|x|}{x}\) Does Not Exist (DNE).

Absolute values are essentially piecewise functions in disguise. When evaluating limits involving them, you MUST consider the definition and often break it down into one-sided limits if the point approaches where the expression inside changes sign. The numerical output from Pyodide clearly shows the different behavior when approaching 0 from the positive and negative sides. This difference is key to why the limit does not exist.

Algebraic Strategy 3: Rationalizing

Resolving Indeterminate Forms with Square Roots

When sums or differences of square roots lead to indeterminate forms (typically \(\frac{0}{0}\) or \(\infty-\infty\)), rationalizing is a powerful technique. This involves multiplying by the conjugate of the expression containing square roots.

Key Identity: Difference of Squares: \(A^2-B^2=(A+B)(A-B)\). Applied to square roots: \((\sqrt{A}-\sqrt{B})(\sqrt{A}+\sqrt{B})=(A)-(B)\).

Process:

1. Identify the indeterminate form, usually \(\frac{0}{0}\).
2. Multiply the numerator and denominator by the conjugate of the expression containing square roots.
3. Simplify the expression, utilizing the difference of squares identity.
4. Look for common factors to cancel.
5. Evaluate the limit of the simplified expression.

Algebraic Strategy 3: Rationalizing

Resolving Indeterminate Forms with Square Roots

Example:

Evaluate \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x+1}-1}{x}\).

1. Direct substitution: \(\frac{\sqrt{0+1}-1}{0} = \frac{1-1}{0} = \frac{0}{0}\). Indeterminate.
2. Multiply by the conjugate of the numerator, \((\sqrt{x+1}+1)\): \[ \frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} \]
3. Simplify the numerator: \((\sqrt{x+1})^2 - (1)^2 = (x+1) - 1 = x\). \[ \frac{x}{x(\sqrt{x+1}+1)} \]
4. Cancel \(x\) (since \(x \ne 0\)): \(\frac{1}{\sqrt{x+1}+1}\).
5. Evaluate the limit: \(\displaystyle\lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{0+1}+1} = \frac{1}{1+1} = \frac{1}{2}\).

Note

“Do More Work” through Rationalization! This technique clears the square roots from the expression, often leading to a cancelable factor.

Rationalizing is a smart algebraic move because it leverages the difference of squares to eliminate the square roots, which are typically the source of the 0/0 problem in these cases. It transforms the expression into a more manageable polynomial or rational form where cancellation can occur. This is a common strategy, especially in problems related to the definition of the derivative for square root functions.

Limits of Piecewise Functions

Checking at the “Break Points”

Limits with piece-wise defined functions operate similarly to those with absolute values. You must consider the specific rule for the function in the region around the limit point.

Key Approach:

When the limit point \(a\) is a “break point” (where the function definition changes):

1. Evaluate the left-hand limit using the function definition valid for \(x < a\).
2. Evaluate the right-hand limit using the function definition valid for \(x > a\).
3. If the left-hand and right-hand limits are equal, then the overall limit exists and is that value. Otherwise, the limit does not exist.

Limits of Piecewise Functions

Checking at the “Break Points”

Example:

Let \(f(x) = \begin{cases} x^2+1, & \text{if } x < 2 \\ 3x-1, & \text{if } x \ge 2 \end{cases}\). Evaluate \(\displaystyle\lim_{x \to 2} f(x)\).

• Left-hand limit: For \(x < 2\), we use \(f(x)=x^2+1\). \(\displaystyle\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2+1) = (2)^2+1 = 5\).
• Right-hand limit: For \(x \ge 2\), we use \(f(x)=3x-1\). \(\displaystyle\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x-1) = 3(2)-1 = 5\).

Since \(\displaystyle\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 5\), the overall limit exists: \(\displaystyle\lim_{x \to 2} f(x) = 5\).

Piecewise functions literally define different functions for different parts of the domain. When you’re approaching a point where the definition changes, you need to carefully check both sides. If the two pieces “meet” at the limit point, the limit exists. If they don’t, there’s a jump discontinuity, and the limit doesn’t exist. This is a direct application of the concept of one-sided limits.

The Squeeze Theorem

Bounding Functions to Find Limits

The Squeeze Theorem (also known as the Sandwich Theorem) is a powerful tool for finding limits of functions that are difficult to evaluate directly, especially those involving oscillatory behavior.

Theorem Statement:

Suppose that \(f(x) \le g(x) \le h(x)\) for all \(x\) that are close to (but not equal to) \(a\).

And suppose that \(\displaystyle\lim_{x \to a} f(x) = L\) and \(\displaystyle\lim_{x \to a} h(x) = L\).

Then, \(\displaystyle\lim_{x \to a} g(x) = L\).

Intuition: If a function \(g(x)\) is always trapped between two other functions, \(f(x)\) and \(h(x)\), and both \(f(x)\) and \(h(x)\) are heading towards the same limit \(L\) as \(x\) approaches \(a\), then \(g(x)\) has no choice but to be “squeezed” to that same limit \(L\).

The Squeeze Theorem

Bounding Functions to Find Limits

graph LR
    A[x --> a] --> B["lim f(x) = L"];
    A --> C["lim h(x) = L"];
    B & C --> D["f(x) <= g(x) <= h(x)"];
    D --> E["Therefore, lim g(x) = L"];

The Squeeze Theorem is perhaps one of the most elegant and useful theorems in beginning calculus. It provides a way to evaluate limits for functions that might not simplify algebraically or yield to direct substitution. Visualizing it as a “squeeze” or “sandwich” helps capture its essence: the function in the middle is forced to go to the same place as the functions on either side.

Squeeze Theorem: A Classic Example

Visualizing the “Squeeze”

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

We know that for any value \(\theta\): \(-1 \le \sin(\theta) \le 1\). So, for \(\theta = 1/x\), we have: \(-1 \le \sin(1/x) \le 1\).

Now, multiply all parts of the inequality by \(x^2\). Since \(x^2 \ge 0\), the inequality signs do not flip:

\(-x^2 \le x^2 \sin(1/x) \le x^2\).

Let \(f(x) = -x^2\), \(g(x) = x^2 \sin(1/x)\), and \(h(x) = x^2\).

We find the limits of \(f(x)\) and \(h(x)\) as \(x \to 0\):

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

Since \(f(x) \le g(x) \le h(x)\) and both \(f(x)\) and \(h(x)\) approach \(0\) as \(x \to 0\), by the Squeeze Theorem, \(\displaystyle\lim_{x \to 0} x^2 \sin(1/x) = 0\).

Important

The Squeeze Theorem is particularly powerful for functions involving \(\sin(1/x)\) or \(\cos(1/x)\) oscillating near a point.

Squeeze Theorem: Interactive Visualization

See the Squeeze in Action!

The oscillating function \(g(x)=x^2\sin(1/x)\) is “squeezed” between \(f(x)=-x^2\) and \(h(x)=x^2\) as \(x\) approaches \(0\).

This interactive plot beautifully illustrates the Squeeze Theorem. Notice how the red x^2 sin(1/x) function rapidly oscillates as it approaches zero, but it cannot escape the bounds of the two parabolas, -x^2 and x^2. As the parabolas converge to zero at the origin, the oscillating function is forced to converge to zero as well. This visual confirmation really drives home the intuition of the theorem.

Key Takeaways: Mastering Limit Computations

Strategies for Success

• Direct Substitution: Always try this first for continuous functions (polynomials, rational functions not dividing by zero).

  Tip

  The simplest path! If it works, you’re done.

• Indeterminate Forms are Signals:

  • \(\frac{\text{non-zero}}{0}\): Implies \(\pm \infty\) or DNE (\(\implies\) vertical asymptote).
  • \(\frac{0}{0}, \infty - \infty\): Require “more work.”
  Warning

  Don’t stop here; these mean there’s more to uncover!

Key Takeaways: Mastering Limit Computations

Strategies for Success

• Algebraic Manipulation is Key:

  • Factoring: For polynomial \(\frac{0}{0}\) forms.
  • Common Denominators: For fractional \(\frac{0}{0}\) or \(\infty - \infty\) forms.
  • Rationalizing (Conjugates): For \(\frac{0}{0}\) forms involving square roots.

• Case Analysis: For functions with absolute values or piecewise definitions, evaluate one-sided limits at “break points.”

• Squeeze Theorem: For oscillatory functions bounded by simpler functions.

  Note

  A powerful tool when algebraic methods fail for bounded, oscillating functions.

To summarize, mastering limits isn’t just about memorizing rules; it’s about developing a strategic approach. Start simple, recognize indeterminate forms, and then apply specific algebraic or geometric techniques. Each “indeterminate form” like 0/0 is an invitation to dig deeper with the tools we’ve covered today. Practice these strategies, and you’ll find that limits become much more approachable. Thank you!