L'Hospital's Rule (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Indeterminate forms

What is an indeterminate form?

An indeterminate form is a mathematical expression of one of the two following forms:
- $\frac{0}{0}$
- $\frac{\pm \infty}{\pm \infty}$
The value of an indeterminate form is undefined
- Dividing by 0 always gives an undefined expression
- And note that, for example, $\frac{\infty}{\infty}$ is not equal to 1
  - $\infty$ is not a number
  - so it can't be canceled to simplify a fraction
Sometimes attempting to evaluate a limit using substitution leads to one of the indeterminate forms given above
- L'Hospital's rule provides a method for dealing with limits of that form

Exam Tip

Other limit methods will also sometimes work when substitution gives an indeterminate form

For example algebraic simplification, multiplying by conjugates or multiplying by reciprocals
- See the 'Evaluating Limits Analytically' study guide

Evaluating limits using L'Hospital's rule

What is L'Hospital’s Rule?

L'Hospital's rule (sometimes written as L’Hôpital’s rule) is a method for finding the value of certain limits using calculus
- Specifically, it allows us to attempt to evaluate the limit of a quotient $\frac{f (x)}{g (x)}$
- for which attempting to evaluate the limit by substitution returns one of the indeterminate forms $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$
For such a quotient function, L'Hospital's rule says that
- $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$
- I.e., you can take the derivatives of the numerator and denominator
  - and attempt to evaluate the limit again in that form

How do I evaluate a limit using L’Hospital’s Rule?

STEP 1
Check that the limit of the quotient results in one of the indeterminate forms given above
- I.e., check that $\lim_{x \to a} \frac{f (x)}{g (x)} = \frac{f (a)}{g (a)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$
STEP 2
Find the derivatives of the numerator and denominator of the quotient
STEP 3
Check whether the limit $\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ exists
STEP 4
If that limit does exist, then $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$
STEP 5
If $\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = \frac{f^{'} (a)}{g^{'} (a)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ then you may repeat the process by considering $\lim_{x \to a} \frac{f^{''} (x)}{g^{''} (x)}$ (and possibly higher order derivatives after that)
- As long as the limits continue giving indeterminate forms you may continue applying L’Hospital’s rule
- Each time this happens find the next set of derivatives and consider the limit again

Exam Tip

Before beginning to use L'Hospital's rule to evaluate a limit

Be sure to confirm that using substitution gives an indeterminate form
Otherwise L'Hospital's rule is not valid

Worked Example

Use L’Hospital’s rule to evaluate each of the following limits:

(a) $\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x}$

Answer:

This limit could also be found by 'multiplying by reciprocals', but the question says to use L'Hospital's rule

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x} = \frac{\infty}{- \infty}$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$\frac{d}{d x} (5 x + 17) = 5 \frac{d}{d x} (4 - 3 x) = - 3$

Apply L'Hospital's Rule, $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x} = \lim_{x \to \infty} \frac{5}{- 3}$

There's no x in that final expression, so x going to infinity doesn't matter!

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x} = - \frac{5}{3}$

(b) $\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x}$

Answer:

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

$\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x} = \frac{{(0)}^{3}}{- 2 (0) + \sin (0)} = \frac{0}{0}$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$\frac{d}{d x} (x^{3}) = 3 x^{2} \frac{d}{d x} (- 2 x + \sin 2 x) = - 2 + 2 \cos 2 x$

Apply L'Hospital's Rule, $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$

$\begin{array}{rcl} \lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x} & = & \lim_{x \to 0} \frac{3 x^{2}}{- 2 + 2 \cos 2 x} \\ = & \frac{3 {(0)}^{2}}{- 2 + 2 \cos (0)} \\ = & \frac{0}{- 2 + 2 (1)} \\ = & \frac{0}{0} \end{array}$

That is another indeterminate form, so we can repeat the process again

$\frac{d}{d x} (3 x^{2}) = 6 x \frac{d}{d x} (- 2 + 2 \cos 2 x) = - 4 \sin 2 x$

Apply L'Hospital's rule again

$\begin{array}{rcl} \lim_{x \to 0} \frac{3 x^{2}}{- 2 + 2 \cos 2 x} & = & \lim_{x \to 0} \frac{6 x}{- 4 \sin 2 x} \\ = & \frac{6 (0)}{- 4 \sin (0)} \\ = & \frac{0}{- 4 (0)} \\ = & \frac{0}{0} \end{array}$

And that is also an indeterminate form, so we can repeat the process again

$\frac{d}{d x} (6 x) = 6 \frac{d}{d x} (- 4 \sin 2 x) = - 8 \cos 2 x$

Apply L'Hospital's rule again

$\begin{array}{rcl} \lim_{x \to 0} \frac{6 x}{- 4 \sin 2 x} & = & \lim_{x \to 0} \frac{6}{- 8 \cos 2 x} \\ = & \frac{6}{- 8 \cos (0)} \\ = & \frac{6}{- 8 (1)} \\ = & - \frac{6}{8} \end{array}$

That is not an indeterminate form, so that's the answer we're looking for

Simplify the fraction

$\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x} = - \frac{3}{4}$

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L'Hospital's Rule (College Board AP® Calculus AB)

Revision Note

Indeterminate forms

What is an indeterminate form?

Exam Tip

Evaluating limits using L'Hospital's rule

What is L'Hospital’s Rule?

How do I evaluate a limit using L’Hospital’s Rule?

Exam Tip

Worked Example

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Author: Roger B