Introduction

This topic introduces L'Hospital's Rule, a powerful technique for evaluating limits that result in indeterminate forms. The essential knowledge tells us that when the ratio of two functions tends to $\frac{0}{0}$ or $\frac{\infty }{\infty }$ in the limit, such forms are said to be indeterminate and limits of the indeterminate forms $\frac{0}{0}$ or $\frac{\infty }{\infty }$ may be evaluated using L'Hospital's Rule. This rule provides an alternative method for finding limits that may be difficult or impossible to evaluate using algebraic techniques.

Understanding Indeterminate Forms

What Are Indeterminate Forms?

An indeterminate form occurs when direct substitution in a limit produces an expression that has no clear meaning.

Two Indeterminate Forms for AP Calculus:

Type 1: $\frac{0}{0}$ - both numerator and denominator approach 0

Type 2: $\frac{\infty }{\infty }$ - both numerator and denominator approach infinity

These forms are "indeterminate" because the limit could be any value, zero, infinity, or may not exist.

Example 1:

Identify the indeterminate form:

$\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}$

Direct substitution: $\frac{4-4}{2-2}=\frac{0}{0}$ (indeterminate form)

Forms That Are NOT Indeterminate

Determinate forms have clear values:

L'Hospital's Rule ONLY applies to $\frac{0}{0}$ and $\frac{\infty }{\infty }$ forms.

L'Hospital's Rule

The Rule Statement

If $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$ produces the indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$, then:

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

provided the limit on the right exists or is $\pm \infty$.

Key Points:

When to Use L'Hospital's Rule

Step 1: Check if direct substitution gives $\frac{0}{0}$ or $\frac{\infty }{\infty }$

Step 2: If yes, apply L'Hospital's Rule

Step 3: Repeat if necessary (the new limit may also be indeterminate)

Solving 0/0 Indeterminate Forms

Polynomial Examples

Example 2:

Evaluate $\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}$

Check the form: $\frac{9-9}{3-3}=\frac{0}{0}$ ✓

Apply L'Hospital's Rule: $\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}=\underset{x\to 3}{\lim }\frac{2x}{1}$

Evaluate: $\frac{2(3)}{1}=6$

Trigonometric Examples

Example 3:

Evaluate $\underset{x\to 0}{\lim }\frac{\sin (x)}{x}$

Check the form: $\frac{\sin (0)}{0}=\frac{0}{0}$ ✓

Apply L'Hospital's Rule: $\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{1}$

Evaluate: $\frac{\cos (0)}{1}=1$

Example 4:

Evaluate $\underset{x\to 0}{\lim }\frac{1-\cos (x)}{x^2}$

Check the form: $\frac{1-1}{0}=\frac{0}{0}$ ✓

Apply L'Hospital's Rule: $\underset{x\to 0}{\lim }\frac{1-\cos (x)}{x^2}=\underset{x\to 0}{\lim }\frac{\sin (x)}{2x}$

Still $\frac{0}{0}$, apply again: $\underset{x\to 0}{\lim }\frac{\sin (x)}{2x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{2}$

Evaluate: $\frac{1}{2}$

Exponential and Logarithmic Examples

Example 5:

Evaluate $\underset{x\to 0}{\lim }\frac{e^x-1}{x}$

Check the form: $\frac{1-1}{0}=\frac{0}{0}$ ✓

Apply L'Hospital's Rule: $\underset{x\to 0}{\lim }\frac{e^x-1}{x}=\underset{x\to 0}{\lim }\frac{e^x}{1}$

Evaluate: $e^0=1$

Solving ∞/∞ Indeterminate Forms

Polynomial Ratios at Infinity

Example 6:

Evaluate $\underset{x\to \infty }{\lim }\frac{3x^2+5x}{2x^2-7}$

Check the form: Both numerator and denominator approach $\infty$ as $x\to \infty$ Form: $\frac{\infty }{\infty }$ ✓

Apply L'Hospital's Rule: $\underset{x\to \infty }{\lim }\frac{3x^2+5x}{2x^2-7}=\underset{x\to \infty }{\lim }\frac{6x+5}{4x}$

Still $\frac{\infty }{\infty }$, apply again: $\underset{x\to \infty }{\lim }\frac{6x+5}{4x}=\underset{x\to \infty }{\lim }\frac{6}{4}$

Evaluate: $\frac{6}{4}=\frac{3}{2}$

Exponential vs. Polynomial Growth

Example 7:

Evaluate $\underset{x\to \infty }{\lim }\frac{x^2}{e^x}$

Check the form: $\frac{\infty }{\infty }$ ✓

Apply L'Hospital's Rule: $\underset{x\to \infty }{\lim }\frac{x^2}{e^x}=\underset{x\to \infty }{\lim }\frac{2x}{e^x}$

Still $\frac{\infty }{\infty }$, apply again: $\underset{x\to \infty }{\lim }\frac{2x}{e^x}=\underset{x\to \infty }{\lim }\frac{2}{e^x}$

Evaluate: $\frac{2}{\infty }=0$

This shows that exponential functions grow faster than polynomials.

Common Mistakes and Cautions

Error 1: Using Quotient Rule

Wrong: Taking the derivative of the entire fraction using quotient rule

Right: Taking derivatives of numerator and denominator separately

Example: For $\frac{x^2}{x}$:

Error 2: Using L'Hospital's When It Doesn't Apply

L'Hospital's Rule ONLY works for $\frac{0}{0}$ and $\frac{\infty }{\infty }$.

Example 8:

Evaluate $\underset{x\to 0}{\lim }\frac{x^2+1}{x}$

Check the form: $\frac{0+1}{0}=\frac{1}{0}$ (NOT indeterminate)

This limit does not exist (approaches $\pm \infty$). Do NOT use L'Hospital's Rule.

Error 3: Forgetting to Re-check

After applying L'Hospital's Rule, always check if the result is still indeterminate before evaluating.

When L'Hospital's Rule May Be Required

Justifying the Use of L'Hospital's Rule

On the AP exam, you should:

Example 9:

Complete justification for $\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}$

The limit has the form $\frac{0}{0}$.

By L'Hospital's Rule: $\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}=\underset{x\to 2}{\lim }\frac{3x^2}{2x}=\frac{12}{4}=3$

Practice Section