Topic 4.7 Notes – Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

1. Indeterminate Forms 0/0 and ∞/∞

When you plug into a limit like

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

sometimes direct substitution gives:

• $\frac{0}{0}$
• $\frac{\infty}{\infty}$

These are called indeterminate forms.

Indeterminate means the expression doesn’t decide the answer. The limit might be:

• A real number
• $\pm \infty$
• Or not exist

For example, near a point $a$, both the numerator and denominator might approach 0 - but one could shrink faster than the other.

Both functions go to 0, but their rates determine the limit of the ratio.

Two quick reminders:

• $0/0$ does not mean the limit is 0.
• $\infty/\infty$ does not mean the limit is 1.
• On AP Calculus AB, these are the only indeterminate forms you’re responsible for.

Before doing anything fancy, always substitute first.

2. L’Hôpital’s Rule

Here’s the rule you’ll use.

If

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \quad \text{or} \quad \frac{\infty}{\infty},$$

then

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

provided the new limit exists.

You differentiate the numerator and denominator separately, then evaluate the new limit.

This is not the quotient rule.
You are not taking the derivative of a fraction.
You are forming a new fraction made of derivatives.

You can apply it:

• As $x \to a$
• As $x \to \infty$
• As $x \to -\infty$

If the new limit is still $0/0$ or $\infty/\infty$, you can apply it again.

3. How to Apply L’Hôpital’s Rule Correctly

Let’s walk through the logic the way graders expect to see it.

Step 1 - Try Direct Substitution

Example:

$$\lim_{x \to 0} \frac{\sin(3x)}{x}$$

Substitute $x = 0$:

• Numerator → 0
• Denominator → 0

We have $0/0$.

Step 2 - Justify the Form (especially on FRQs)

You’d show:

$$\lim_{x \to 0} \sin(3x) = 0$$

$$\lim_{x \to 0} x = 0$$

That confirms L’Hôpital applies.

Step 3 - Differentiate Top and Bottom

$$\frac{\sin(3x)}{x} \rightarrow \frac{3\cos(3x)}{1}$$

(Chain rule on the numerator.)

Step 4 - Re-evaluate the Limit

$$\lim_{x \to 0} 3\cos(3x) = 3$$

So the original limit equals 3.

Notice something important. The limit became straightforward after differentiating once. That’s typical.

4. When to Use It and When Not To

L’Hôpital is helpful when:

• Trig expressions give 0/0
• Rational functions give $\infty/\infty$
• Exponentials and polynomials compete at infinity
• Algebraic simplification is messy or impossible

Example at infinity:

$$\lim_{x \to \infty} \frac{5x^3 - 2}{9x^3 + x}$$

Substitution gives $\infty/\infty$.
Differentiate:

$$\frac{15x^2}{27x^2 + 1}$$

Now take the limit → $\frac{15}{27} = \frac{5}{9}$.

That matches what you’d get by dividing by highest power, so sometimes algebra is quicker.

Avoid L’Hôpital when:

• Direct substitution gives a number.
• The limit isn’t written as a fraction.
• Factoring or canceling solves it instantly.
• The form isn’t $0/0$ or $\infty/\infty$.

Students often lose time by jumping to L’Hôpital when basic algebra works faster.

5. Common Mistakes That Cost Points

• Not verifying the indeterminate form first. On free-response, you need to show it.
• Using it when the form isn’t 0/0 or ∞/∞.
• Forgetting to take the limit after differentiating.
• Stopping when it’s still indeterminate. You can apply the rule again.
• Derivative mistakes. Most wrong answers here come from chain rule or trig errors.

On multiple choice, they love answer choices that match common derivative mistakes.