4.7 - Using L'Hospital's Rule for Determining Limits of Indeterminate Forms

Historical Background

Guillaume de l’Hôpital (1661 to 1704) was a French mathematician who published one of the first calculus textbooks. The rule named after him gives us a powerful way to evaluate limits that produce indeterminate forms.

When to Use L’Hôpital’s Rule

L’Hôpital’s Rule applies only when direct substitution produces one of these indeterminate forms:

\frac{0}{0} \quad \text{or} \quad \frac{\infty}{\infty}

There are other indeterminate forms (like

\infty - \infty

and

0 \cdot \infty

), but the AP exam only tests

\frac{0}{0}

and

\frac{\infty}{\infty}

Always Start with Direct Substitution

Before you reach for any special technique, try plugging in the limit value directly.

Example: Evaluate

\displaystyle\lim_{x \to 4} \frac{x + 1}{2x - 3}

Direct substitution gives

\frac{4 + 1}{2(4) - 3} = \frac{5}{5} = 1

The limit equals

1

, and no special technique was needed.

L’Hôpital’s Rule Statement

\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)}

produces

\frac{0}{0}

\frac{\infty}{\infty}

, then:

\boxed{\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}}

In words: take the derivative of the numerator, take the derivative of the denominator, then evaluate the limit.

Important Notes

Differentiate the numerator and denominator separately

This is NOT the quotient rule

It only works for limits producing $\frac{0}{0}$ or $\frac{\infty}{\infty}$

You may need to apply it multiple times

It needs $f$ and $g$ to be differentiable near $a$ , and it gives the answer when the new limit $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists (or is $\infty$ or $-\infty$ )

Example 1: Basic Application ( $\frac{0}{0}$ Form)

Evaluate

\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 + x - 12}

Step 1: Check with direct substitution

\frac{3^2 - 9}{3^2 + 3 - 12} = \frac{9 - 9}{9 + 3 - 12} = \frac{0}{0}

Indeterminate form, so we can use L’Hôpital’s Rule. ✓

Step 2: Apply L’Hôpital’s Rule

Take the derivatives of the numerator and denominator separately:

\lim_{x \to 3} \frac{x^2 - 9}{x^2 + x - 12} = \lim_{x \to 3} \frac{2x}{2x + 1}

Step 3: Evaluate

= \frac{2(3)}{2(3) + 1} = \frac{6}{7}

Answer:

\frac{6}{7}

Example 2: Polynomial Ratio

Evaluate

\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x^4 - 2x + 1}

Step 1: Check the indeterminate form

\frac{1^3 - 1}{1^4 - 2(1) + 1} = \frac{0}{0} \quad \checkmark

Step 2: Apply L’Hôpital’s Rule

\lim_{x \to 1} \frac{x^3 - 1}{x^4 - 2x + 1} = \lim_{x \to 1} \frac{3x^2}{4x^3 - 2}

Step 3: Evaluate

= \frac{3(1)^2}{4(1)^3 - 2} = \frac{3}{4 - 2} = \frac{3}{2}

Answer:

\frac{3}{2}

Example 3: Trigonometric Limits

Evaluate

\displaystyle\lim_{x \to 0} \frac{\sin(4x)}{x}

Note: you might recognize this as a special trig limit, but L’Hôpital’s Rule works too.

Step 1: Check the indeterminate form

\frac{\sin(0)}{0} = \frac{0}{0} \quad \checkmark

Step 2: Apply L’Hôpital’s Rule

\lim_{x \to 0} \frac{\sin(4x)}{x} = \lim_{x \to 0} \frac{4\cos(4x)}{1}

Remember to use the chain rule here:

\frac{d}{dx}[\sin(4x)] = \cos(4x) \cdot 4

Step 3: Evaluate

= \frac{4\cos(0)}{1} = \frac{4(1)}{1} = 4

Answer:

4

Example 4: Applying L’Hôpital’s Rule Multiple Times

Evaluate

\displaystyle\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}

Step 1: Check the indeterminate form

\frac{1 - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0} \quad \checkmark

Step 2: Apply L’Hôpital’s Rule, first time

\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{\sin(x)}{2x}

Step 3: Check again

\frac{\sin(0)}{2(0)} = \frac{0}{0}

Still indeterminate, so apply L’Hôpital’s Rule again.

Step 4: Apply L’Hôpital’s Rule, second time

\lim_{x \to 0} \frac{\sin(x)}{2x} = \lim_{x \to 0} \frac{\cos(x)}{2}

Step 5: Evaluate

= \frac{\cos(0)}{2} = \frac{1}{2}

Answer:

\frac{1}{2}

Key lesson: if you still get

\frac{0}{0}

\frac{\infty}{\infty}

after applying L’Hôpital’s once, apply it again, and again if you need to.

Example 5: Limits at Infinity ( $\frac{\infty}{\infty}$ Form)

Evaluate

\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5}{e^x}

Step 1: Check the indeterminate form

x \to \infty

, the numerator goes to

\infty

and the denominator goes to

\infty

\frac{\infty}{\infty} \quad \checkmark

Step 2: Apply L’Hôpital’s Rule, first time

\lim_{x \to \infty} \frac{3x^2 + 5}{e^x} = \lim_{x \to \infty} \frac{6x}{e^x}

Step 3: Check again

Still

\frac{\infty}{\infty}

, so apply it again.

Step 4: Apply L’Hôpital’s Rule, second time

\lim_{x \to \infty} \frac{6x}{e^x} = \lim_{x \to \infty} \frac{6}{e^x}

Step 5: Evaluate

= \frac{6}{e^\infty} = \frac{6}{\infty} = 0

Answer:

0

Interpretation: exponential functions grow faster than polynomials, so the denominator dominates.

Example 6: Another Infinity Limit

Evaluate

\displaystyle\lim_{x \to \infty} \frac{5x^3}{2x^3 + x}

Step 1: Check the indeterminate form

\frac{\infty}{\infty} \quad \checkmark

Step 2: Apply L’Hôpital’s Rule

\lim_{x \to \infty} \frac{5x^3}{2x^3 + x} = \lim_{x \to \infty} \frac{15x^2}{6x^2 + 1}

Step 3: Check again

Still

\frac{\infty}{\infty}

, so apply it again.

Step 4: Apply L’Hôpital’s Rule, second time

\lim_{x \to \infty} \frac{15x^2}{6x^2 + 1} = \lim_{x \to \infty} \frac{30x}{12x}

Step 5: Simplify and evaluate

= \lim_{x \to \infty} \frac{30x}{12x} = \lim_{x \to \infty} \frac{30}{12} = \frac{5}{2}

Answer:

\frac{5}{2}

Common Mistake: L’Hôpital’s Rule vs. the Quotient Rule

This is L’Hôpital’s Rule, for limits:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{(when appropriate)}

This is the Quotient Rule, for derivatives:

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Example: Taking a Derivative (NOT L’Hôpital’s)

Find

\frac{d}{dx}\left[\frac{\sin(3x)}{x}\right]

This is asking for a derivative, not a limit, so use the quotient rule:

\frac{d}{dx}\left[\frac{\sin(3x)}{x}\right] = \frac{3\cos(3x) \cdot x - \sin(3x) \cdot 1}{x^2} = \frac{3x\cos(3x) - \sin(3x)}{x^2}

Do NOT just differentiate the top and bottom separately. That’s only for L’Hôpital’s Rule with limits.

When L’Hôpital’s Rule Does NOT Apply

You cannot use L’Hôpital’s Rule if direct substitution gives:

$\frac{5}{0}$ , which is undefined, not indeterminate

$\frac{0}{5}$ , which equals zero, already determined

$\frac{7}{3}$ , already determined

$\frac{\infty}{5}$ , which equals infinity, already determined

Example: When NOT to Use It

Evaluate

\displaystyle\lim_{x \to 2} \frac{x^2 + 1}{x - 2}

Direct substitution gives

\frac{2^2 + 1}{2 - 2} = \frac{5}{0}

That is not an indeterminate form. The limit does not exist, since it approaches

\pm\infty

depending on the direction.

Do not use L’Hôpital’s Rule here.

Showing Your Work on the AP Exam

On a free-response question, you have to show that L’Hôpital’s Rule applies before you use it. Write the limits of the numerator and the denominator separately:

“Since

\displaystyle\lim_{x \to 3} (x^2 - 9) = 0

and

\displaystyle\lim_{x \to 3} (x^2 + x - 12) = 0

, L’Hôpital’s Rule applies.”

Then write the new limit and evaluate it, like in Example 1.

Avoid writing

\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 + x - 12} = \frac{0}{0}

\frac{0}{0}

isn’t a number, so a limit can’t equal it. Show the two separate limits instead.

Strategy Summary

Try direct substitution

If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , apply L’Hôpital’s Rule

Differentiate the numerator and denominator separately

Evaluate the new limit

If it’s still indeterminate, repeat L’Hôpital’s Rule

Key Takeaways

L’Hôpital’s Rule: $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ when you have $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Always check first. Use direct substitution to confirm an indeterminate form

Show the conditions. On free-response questions, state that the numerator and denominator both approach 0 (or both approach $\pm\infty$ ) before using the rule

Differentiate separately. Take the derivative of the numerator and the denominator independently

It’s not the quotient rule. L’Hôpital’s is for limits only, not general derivatives

You may need multiple applications. If it’s still $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , apply it again

Only two forms. L’Hôpital’s Rule only applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$

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4.7 - Using L'Hospital's Rule for Determining Limits of Indeterminate Forms

Historical Background

When to Use L’Hôpital’s Rule

Always Start with Direct Substitution

L’Hôpital’s Rule Statement

Important Notes

Example 1: Basic Application (00\frac{0}{0}00​ Form)

Example 2: Polynomial Ratio

Example 3: Trigonometric Limits

Example 4: Applying L’Hôpital’s Rule Multiple Times

Example 5: Limits at Infinity (∞∞\frac{\infty}{\infty}∞∞​ Form)

Example 6: Another Infinity Limit

Common Mistake: L’Hôpital’s Rule vs. the Quotient Rule

This is L’Hôpital’s Rule, for limits:

This is the Quotient Rule, for derivatives:

Example: Taking a Derivative (NOT L’Hôpital’s)

When L’Hôpital’s Rule Does NOT Apply

Example: When NOT to Use It

Showing Your Work on the AP Exam

Strategy Summary

Key Takeaways

Example 1: Basic Application ( $\frac{0}{0}$ Form)

Example 5: Limits at Infinity ( $\frac{\infty}{\infty}$ Form)