# L'Hôpital's rule (HL only)

> IB Mathematics: Analysis and Approaches HL · IB AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-l-h-pital-s-rule/

This module explains how to use L'Hôpital's rule to evaluate indeterminate limits, including 0/0, ∞/∞, 0×∞, ∞−∞, and indeterminate power forms. You will learn the rule's conditions and how to avoid common exam errors.

**Prerequisites:** [Limits and continuity](https://www.owlsprep.com/study/ib-math-aa-hl-u5-limits-and-continuity/); [Rules of differentiation](https://www.owlsprep.com/study/ib-math-aa-hl-u4-rules-of-differentiation/)

## Learning objectives

- State the conditions required for L'Hôpital's rule to apply
- Evaluate indeterminate limits of the form 0/0 and ∞/∞ using L'Hôpital's rule
- Adapt L'Hôpital's rule for other indeterminate forms like 0×∞, ∞−∞ and indeterminate powers
- Identify common mistakes when applying the rule

## Conditions and Core Statement

**L'Hôpital's Rule** — If: (1) $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or both $\pm\infty$, (2) $f$ and $g$ are differentiable near $a$, (3) $g'(x) \neq 0$ near $a$, (4) $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$. The rule works for one-sided limits and limits at infinity.

*Notation:* For $\boxed{\lim_{x \to a} \frac{f(x)}{g(x)}}$

*Example:* Applies to $\lim_{x \to 0} \frac{\sin x}{x}$, which is 0/0 indeterminate.

The most critical step is confirming the original limit is indeterminate. If it is not indeterminate, applying the rule will always give an incorrect result.

**Worked example:** Verify that L'Hôpital's rule can be applied to $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$, then evaluate the limit.

1. Check the indeterminate form by substituting $x=2$:
2. $$f(2) = 2^2 - 4 = 0, \quad g(2) = 2 - 2 = 0$$
3. We have a 0/0 indeterminate form. Both are polynomials, so differentiable everywhere, and $g'(x) = 1 \neq 0$ for all $x$. All conditions are satisfied.
4. Differentiate numerator and denominator separately:
5. $$f'(x) = 2x, \quad g'(x) = 1$$
6. Evaluate the limit of the ratio of derivatives:
7. $$\lim_{x \to 2} \frac{2x}{1} = 4$$
8. By L'Hôpital's rule, the original limit equals 4.

> **tip**
>
> Always confirm the original limit is indeterminate before writing you are using L'Hôpital's rule; exam markers deduct marks for skipping this step.

## Basic Indeterminate Forms: 0/0 and ∞/∞

If after one application of L'Hôpital's rule the new ratio is still indeterminate, you can reapply the rule as long as all conditions still hold.

**Worked example:** Evaluate $\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{5x^2 + 4}$.

1. Check the form as $x \to \infty$: both numerator and denominator approach $+\infty$, so we have an ∞/∞ indeterminate form, all conditions are satisfied.
2. Apply L'Hôpital's rule:
3. $$\lim_{x \to \infty} \frac{6x - 2}{10x}$$
4. This is still ∞/∞ indeterminate, so apply the rule again:
5. $$\lim_{x \to \infty} \frac{6}{10} = \frac{3}{5}$$
6. The original limit is $\frac{3}{5}$, matching the result from dividing numerator and denominator by $x^2$.

**Check your understanding**

Test your understanding

1. Which of the following limits can L'Hôpital's rule be applied to immediately?

   - $\lim_{x \to 0} \frac{x}{x + 1}$
   - $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$
   - $\lim_{x \to 1} \frac{x^2 + 1}{x + 1}$

   *Answer:* $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

   *Why:* Correct. $\lim_{x \to 0} (1-\cos x) = 0$ and $\lim_{x \to 0} x^2 = 0$, so this is 0/0 indeterminate.

## Other Indeterminate Forms: 0×∞ and ∞−∞

L'Hôpital's rule is only stated for ratios of functions, but we can rearrange other indeterminate forms into 0/0 or ∞/∞ to apply the rule.

For $0 \times \infty$, rewrite the product as a ratio by moving one term to the denominator: $f(x)g(x) = \frac{f(x)}{1/g(x)}$, which will become 0/0 or ∞/∞. For $\infty - \infty$, combine terms into a single fraction, usually via a common denominator.

**Worked example:** Evaluate $\lim_{x \to 0^+} x \ln x$.

1. Check the form: as $x \to 0^+$, $x \to 0$ and $\ln x \to -\infty$, so we have a $0 \times (-\infty)$ indeterminate form.
2. Rewrite as a ratio to get ∞/∞:
3. $$x \ln x = \frac{\ln x}{1/x}$$
4. Now $\ln x \to -\infty$ and $1/x \to +\infty$, so we have ∞/∞. All conditions are satisfied. Apply L'Hôpital's rule:
5. Differentiate numerator and denominator:
6. $$\frac{d}{dx} \ln x = \frac{1}{x}, \quad \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2}$$
7. Simplify the ratio of derivatives:
8. $$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$
9. So $\lim_{x \to 0^+} x \ln x = 0$.

> **tip**
>
> When rewriting $0 \times \infty$, put the term that becomes simpler after differentiation in the numerator. Putting the wrong term in the numerator leads to more complicated derivatives that do not help.

## Indeterminate Powers: $0^0, \infty^0, 1^\infty$

For indeterminate powers of the form $\lim f(x)^{g(x)}$, use natural logarithm to convert the power into a product, which can then be rearranged for L'Hôpital's rule. The standard process is:

1. Let $y = f(x)^{g(x)}$
2. Take natural logs: $\ln y = g(x) \ln f(x)$
3. Evaluate $\lim \ln y$ using the 0×∞ method
4. Exponentiate to get $\lim y = e^{\lim \ln y}$

**Worked example:** Evaluate $\lim_{x \to 0^+} x^x$.

1. Let $y = x^x$, take natural logs of both sides:
2. $$\ln y = x \ln x$$
3. We already found that $\lim_{x \to 0^+} x \ln x = 0$ from the previous example, so:
4. $$\lim_{x \to 0^+} \ln y = 0$$
5. Exponentiate both sides to get the original limit:
6. $$\lim_{x \to 0^+} y = \lim_{x \to 0^+} e^{\ln y} = e^0 = 1$$
7. So $\lim_{x \to 0^+} x^x = 1$.

## Common pitfalls

- **Wrong:** Applying L'Hôpital's rule to a non-indeterminate limit, e.g. $\lim_{x \to 0} \frac{x}{x+1}$.
  - Why it fails: The rule only applies when the original limit is 0/0 or ∞/∞. A non-indeterminate limit can be evaluated directly.
  - Correct: Check that both numerator and denominator approach 0 or both approach ±∞ before applying the rule.
- **Wrong:** Differentiating the entire fraction using the quotient rule, instead of differentiating numerator and denominator separately.
  - Why it fails: L'Hôpital's rule requires the ratio of individual derivatives, not the derivative of the ratio.
  - Correct: Always write $\frac{f'(x)}{g'(x)}$, not $\left(\frac{f}{g}\right)'(x)$.
- **Wrong:** Stopping after one application when the derivative ratio is still indeterminate.
  - Why it fails: If the first ratio is still indeterminate, you need to reapply the rule to get the correct result.
  - Correct: Check the form of the derivative ratio after each application, and reapply if it is still indeterminate.
- **Wrong:** Applying the rule directly to 0×∞, ∞−∞, or indeterminate powers without rearrangement.
  - Why it fails: L'Hôpital's rule is only defined for ratios of functions, so applying it directly to other forms is invalid.
  - Correct: Always rearrange the expression into a 0/0 or ∞/∞ ratio before applying L'Hôpital's rule.

## Cheatsheet

| Indeterminate Form | How to apply L'Hôpital's rule |
| --- | --- |
| 0/0, ∞/∞ | Apply directly after checking conditions |
| 0 × ∞ | Rewrite as $\frac{f}{1/g}$ to get 0/0 or ∞/∞ |
| ∞ − ∞ | Combine into a single fraction to get 0/0 or ∞/∞ |
| $0^0, \infty^0, 1^\infty$ | Take natural log to convert to 0 × ∞ |

## What's next

L'Hôpital's rule is a core HL tool for evaluating indeterminate limits, which are required for improper integrals, testing series convergence, and finding limits of sequences in IB AA HL. It is commonly tested alongside differentiation and series topics, and confirming the rule's conditions will help you avoid losing easy marks in exams. Mastery of this method will make more advanced calculus topics significantly easier to work through.

- [Definition of the derivative](https://www.owlsprep.com/study/ib-math-aa-hl-u5-definition-of-the-derivative/)
- [Differentiation rules](https://www.owlsprep.com/study/ib-math-aa-hl-u5-differentiation-rules/)

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