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

Note that if substitution gives limits that look like $\frac{0}{\pm \infty }$ or $\frac{\pm \infty }{0}$, these are not indeterminate forms, and L'Hospital's rule cannot be used

• In the first case, $\frac{0}{\pm \infty }$, the limit will just be equal to $0$

• In the second case, $\frac{\pm \infty }{0}$, the limit will diverge to either $+\infty$ or $-\infty$ depending on the behavior near the limit point

  • See the 'Infinite Limits & Limits at Infinity' study guide

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

Notice that the formula includes the ratio of the derivatives. This is different to $\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)$. If you are using the quotient rule, then it is likely that you have made a mistake.

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

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

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

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

Answer:

(a)

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

$\underset{x\to \infty }{\lim }\left(5x+17\right)\to \infty$ and $\underset{x\to \infty }{\lim }\left(4-3x\right)\to -\infty$

$\underset{x\to \infty }{\lim }\frac{5x+17}{4-3x}$ leads to $\frac{\infty }{-\infty }$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$$\begin{gathered} \frac{d}{dx}\left(5x+17\right)=5 \\ \frac{d}{dx}\left(4-3x\right)=-3 \end{gathered}$$

Apply L'Hospital's Rule, $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\text{'}}\left(x\right)}{g^{\text{'}}\left(x\right)}$

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

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

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

(b)

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

$\underset{x\to 0}{\lim }{x}^3=0^3=0$

$\underset{x\to 0}{\lim }\left(-2x+\sin 2x\right)=-2·0+\sin \left(2·0\right)=0$

$\underset{x\to 0}{\lim }\frac{x^3}{-2x+\sin 2x}$ leads to $\frac{0}{0}$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$$\begin{gathered} \frac{d}{dx}\left(x^3\right)=3x^2 \\ \frac{d}{dx}\left(-2x+\sin 2x\right)=-2+2\cos 2x \end{gathered}$$

Apply L'Hospital's Rule, $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\text{'}}\left(x\right)}{g^{\text{'}}\left(x\right)}$

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{x^3}{-2x+\sin 2x}& =& \underset{x\to 0}{\lim }\frac{3x^2}{-2+2\cos 2x}\end{array}$

Show that this is another indeterminate form

$\underset{x\to 0}{\lim }3x^2=3·0^2=0$

$\underset{x\to 0}{\lim }\left(-2+2\cos 2x\right)=-2+2\cos \left(2·0\right)=0$

This leads to $\begin{array}{rcl}& & \frac{0}{0}\end{array}$ again

Apply L'Hospital's rule again

$$\begin{gathered} \frac{d}{dx}\left(3x^2\right)=6x \\ \frac{d}{dx}\left(-2+2\cos 2x\right)=-4\sin 2x \end{gathered}$$

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

Show that this is also an indeterminate form

$\underset{x\to 0}{\lim }6x=6·0=0$

$\underset{x\to 0}{\lim }\left(-4\sin 2x\right)=-4\sin \left(2·0\right)=0$

This leads to $\begin{array}{rcl}& & \frac{0}{0}\end{array}$ again

Apply L'Hospital's rule again

$$\begin{gathered} \frac{d}{dx}\left(6x\right)=6 \\ \frac{d}{dx}\left(-4\sin 2x\right)=-8\cos 2x \end{gathered}$$

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{6x}{-4\sin 2x}& =& \underset{x\to 0}{\lim }\frac{6}{-8\cos 2x}\end{array}$

Check that this is not an indeterminate form

$\underset{x\to 0}{\lim }6=6$

$\underset{x\to 0}{\lim }\left(-8\cos 2x\right)=-8\cos \left(2·0\right)=-8$

Find the limit by substitution

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{6}{-8\cos 2x}& =& -\frac{6}{8}\end{array}$

Simplify the fraction

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