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

1. Indeterminate Forms 0/0 and ∞/∞

Sometimes you plug a value into

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

and get something that doesn’t tell you the answer.

Two cases you’re responsible for on the AP exam:

• $0/0$ → numerator → 0 and denominator → 0
• $\infty/\infty$ → numerator and denominator both grow without bound (or both → −∞)

These are called indeterminate forms because they don’t determine a single outcome. Different functions can give different limits even if they both produce $0/0$.

Example idea:

• $\frac{x}{x} \to 1$ as $x\to0$
• $\frac{x^2}{x} \to 0$ as $x\to0$

Both give $0/0$ at first glance, but different answers.

Important scope note: other indeterminate forms like $0\cdot\infty$, $1^\infty$, etc., are not required for the AP Calculus exam.

Before using L’Hôpital’s Rule, you must verify:

• $\lim_{x\to a} f(x)=0$ and $\lim_{x\to a} g(x)=0$, or
• both limits are $\pm\infty$

You’re checking the limits, not just plugging in casually.

2. L’Hôpital’s Rule

If

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

produces $0/0$ or $\infty/\infty$, and $f$ and $g$ are differentiable near $a$, then

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

as long as the new limit exists (or is ±∞).

What you actually do:

• Differentiate the top
• Differentiate the bottom
• Keep it as a fraction
• Re-evaluate the limit

This is not the quotient rule. You are not finding the derivative of the whole fraction. You’re forming a new fraction of derivatives.

You can use it for:

• Limits as $x\to a$
• One-sided limits
• Limits as $x\to \infty$ or $-\infty$

3. How to Apply L’Hôpital’s Rule

Here’s the clean process you should show on an FRQ.

1. Substitute and check the form.
   Suppose

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

   Substitution gives $0/0$.

2. Justify the indeterminate form.
   $\lim_{x\to0} \sin(3x)=0$
   $\lim_{x\to0} x=0$

3. Differentiate numerator and denominator.

   $$\frac{d}{dx}[\sin(3x)] = 3\cos(3x), \quad \frac{d}{dx}[x] = 1$$

   New limit:

   $$\lim_{x\to0} \frac{3\cos(3x)}{1}$$

4. Evaluate again.
   $3\cos(0)=3$

So the original limit equals 3.

If you still get $0/0$ or $\infty/\infty$, you can apply L’Hôpital again. That’s common with higher powers.

4. Limits at Infinity

L’Hôpital is often used for polynomial ratios.

Example:

$$\lim_{x\to\infty} \frac{5x^3 - 2x}{2x^3 + 7}$$

Substitution gives $\infty/\infty$.

Differentiate top and bottom:

$$\lim_{x\to\infty} \frac{15x^2 - 2}{6x^2}$$

Still $\infty/\infty$, so apply again:

$$\lim_{x\to\infty} \frac{30x}{12x}$$

Still $\infty/\infty$, apply again:

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

Notice how repeated differentiation strips away powers until constants remain.

That said, for pure polynomials, dividing by the highest power is often faster on a no-calculator multiple choice question. Choose the efficient method.

5. When Not to Use It

L’Hôpital only works when:

• The expression is a quotient
• The form is $0/0$ or $\infty/\infty$

Do not use it if:

• You get a nonzero number over 0 → that’s a vertical asymptote.
• The limit is easy by factoring or canceling.
• The expression is not written as a fraction.

A common mistake on quizzes is applying L’Hôpital automatically whenever you see a fraction. Always check the form first.