3.6 L’Hopital’s Rule

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle }$

In this section we compute limits using L’Hopital’s Rule which requires our knowledge of derivatives.

1 L’Hopital’s Rule

L’Hopital’s Rule uses the derivative to help us find limits involving indeterminate forms. The main indeterminate forms we will discuss are $\frac 00, \frac {\infty }{\infty }, 0\cdot \infty , 1^\infty$ and $0^0$ . We begin with the fractional forms.

L’Hopital’s Rule

$\text {If} \;\; \lim _{x \to c} \frac {f(x)}{g(x)} = \frac {0}{0} \; \text {or} \; \frac {\pm \infty }{\pm \infty }, \;\;\text {then} \; \lim _{x \to c} \frac {f(x)}{g(x)} = \lim _{x \to c} \frac {f'(x)}{g'(x)}$ provided the latter limit exists. The statement is also true for one-sided limits and if ‘ $c$ ’ is replaced by $\infty$ or $-\infty$ .

The indeterminate form $\frac 00$

example 1 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {\sin (3x)}{6x}}.$

Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies.

We have

$\lim _{x \to 0} \frac {\sin (3x)}{6x} = \lim _{x \to 0} \frac {3\cos (3x)}{6} = \frac 12$

(problem 1a) Compute $\lim _{x \to 0} \frac {1 - \cos (x)}{x} = \answer {0}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 1b) Compute $\lim _{x \to 0} \frac {\tan (x)}{3x} = \answer {1/3}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\tan (x)$ is $\sec ^2(x)$

Compute the limit of the new fraction

(problem 1c)

Let $m$ and $m$ be non-zero constants. Compute

$\lim _{x \to 0} \frac {\sin (mx)}{nx} = \answer {m/n}$

Check for a fractional indeterminate form.

Use the chain rule in the numerator.

(problem 1d)

Compute

$\lim _{x \to 0} \frac {\sin ^2(x)}{3x} = \answer {0}$

Check for a fractional indeterminate form.

Use the chain rule in the numerator.

example 2 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {e^{4x} - 1}{12x}}.$
Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {e^{4x} - 1}{12x} = \lim _{x \to 0} \frac {4e^{4x}}{12} = \frac 13$

(problem 2a) Compute $\lim _{x \to 0} \frac {5x}{2 - 2e^x} = \answer {-5/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 2b) Compute $\lim _{x \to 0} \frac {e^{2x} -1}{7\sin (x)} = \answer {2/7}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $e^{2x}$ is $2e^{2x}$ , by the Chain Rule

Compute the limit of the new fraction

(problem 2c) Let $a$ and $b$ be non-zero constants. Compute $\lim _{x \to 0} \frac {e^{ax} - 1}{bx} = \answer {a/b}$

Check for fractional indeterminate form.

Use the chain rule in the numerator.

example 3 Compute $\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32}.$

Plugging in the terminal value, $x=2$ , gives the indeterminate form $\frac 00$ , so we can use L’Hopital’s Rule:

$\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32} = \lim _{x \to 2} \frac {3x^2}{5x^4} = \frac {12}{80} = \frac {3}{20}.$

(problem 3a) Compute $\lim _{x \to 3} \frac {x^3 - 27}{x^4 - 81} = \answer {1/4}$

Plug in $x=3$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 3b) Compute $\lim _{x \to 4} \frac {x^2 - 3x - 4}{x^3 - 64} = \answer {5/48}$

Plug in $x=4$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 3c) Let $m$ and $n$ be non-zero constants. Compute $\lim _{x \to a} \frac {x^m - a^m}{x^n - a^n} = \answer {m/n a^{m-n}}$

Check for fractional indeterminate form.

Use the power rule in the numerator and denominator.

The derivative of a constant is zero.

Sometimes we have to use L’Hopital’s Rule more than once.

example 4 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {e^x - x - 1}{x^2}}.$

Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {e^x - x - 1}{x^2} = \lim _{x \to 0} \frac {e^x -1}{2x} = \frac {0}{0}.$ Applying L’Hopital’s Rule again gives $\lim _{x \to 0} \frac {e^x -1}{2x} = \frac {e^x}{2} = \frac {1}{2}.$ Hence $\displaystyle {\lim _{x \to 0} \frac {e^x - x - 1}{x^2} = \frac 12}.$

(problem 4a) Compute $\lim _{x \to 0} \frac {1 - \cos (x)}{x^2} = \answer {1/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

If you get $\frac 00$ , use L’Hopital’s Rule again

(problem 4b) Compute $\lim _{x \to 0} \frac {\sec (x) - 1}{x^3 - x^2} = \answer {-1/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\sec (x)$ is $\sec (x)\tan (x)$

Compute the limit of the new fraction

If you get $\frac 00$ , use L’Hopital’s Rule again

Use the Product Rule to find the derivative of $\sec (x)\tan (x)$

example 5 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {\tan (x)-x}{x^3}}.$
Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {\tan (x)-x}{x^3} = \lim _{x \to 0} \frac {\sec ^2(x)-1}{3x^2} = \frac {0}{0}.$ Applying L’Hopital’s Rule again gives $\lim _{x \to 0} \frac {\sec ^2(x)-1}{3x^2} = \lim _{x \to 0} \frac {2\sec ^2(x)\tan (x)}{6x} = \frac {0}{0}.$ We need to apply L’Hopital’s Rule again, but first, the numerator is complicated and $\lim _{x \to 0} \sec (x) = 1 \neq 0$ so we take a simplifying step before applying the rule. $\lim _{x \to 0} \frac {2\sec ^2(x)\tan (x)}{6x} = \lim _{x \to 0} \sec ^2(x) \cdot \lim _{x \to 0} \frac {2\tan (x)}{6x}=$ $1\cdot \lim _{x \to 0} \frac {2\sec ^2(x)}{6} =\frac {1}{3}.$

We next consider problems of the form $\frac {\infty }{\infty }$ . These are handled the same way as the $\frac 00$ case above.

The $\dfrac {\infty }{\infty }$ case.

example 6 Compute the limit: $\displaystyle {\;\lim _{x \to \infty } \frac {x^2}{e^x}}.$

As $x$ approaches $\infty$ we get the indeterminate form $\frac {\infty }{\infty }$ so L’Hopital’s Rule applies. We have $\lim _{x\to \infty } \frac {x^2}{e^x} = \lim _{x \to \infty } \frac {2x}{e^x} = \frac {\infty }{\infty }.$ Applying L’Hopital again, we get $\lim _{x \to \infty } \frac {2x}{e^x} = \lim _{\infty } \frac {2}{e^x} = 0.$ Hence $\lim _{x\to \infty }\frac {x^2}{e^x}=0$ . This limit can be generalized as follows: $\lim _{x\to \infty }\frac {x^n}{e^x} =0$ for any exponent $n = 1, 2, 3, \dots$ . This general result comes from using L’Hopital’s Rule $n$ times, yielding $\lim _{x\to \infty } \frac {n!}{e^x} =0$ where $n! = n(n-1)(n-2)\cdot \dots \cdot 3\cdot 2\cdot 1$ . The interpretation of this limit is that the exponential function $e^x$ grows faster than any power of $x$ as $x \to \infty$ .

(problem 6a) Compute $\lim _{x \to \infty } \frac {3x+2}{2x + 3} = \answer {3/2}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 6b) Compute $\lim _{x \to \infty } \frac {5x^2 + 2x - 3}{4x^2 - x + 1} = \answer {5/4}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule again

example 7 Compute the limit: $\displaystyle {\;\lim _{x \to \infty } \frac {\sqrt {x}}{\ln (x)}}$ . As $x \to \infty$ we get $\frac {\infty }{\infty }$ , so L’Hopital’s Rule applies. We have: $\lim _{x \to \infty } \frac {\sqrt x}{\ln (x)} = \lim _{x \to \infty }\frac {\left (\frac {1}{2\sqrt x}\right )}{\left (\frac {1}{x}\right )}$ which simplifies to $\lim _{x\to \infty } \frac {x}{2\sqrt x} = \lim _{x\to \infty }\frac {\sqrt x}{2} = \infty .$ Hence, $\lim _{x\to \infty } \frac {\sqrt x}{\ln (x)} = \infty$ . The interpretation of this limit is that $\sqrt x$ goes to $\infty$ faster than $\ln (x)$ as $x\to \infty$ .

(problem 7a) Compute $\lim _{x \to \infty } \frac {\sqrt x}{2x + 3} = \answer {0}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\sqrt x$ is $\frac {1}{2\sqrt x}$

Compute the limit of the new fraction

(problem 7b) Compute $\lim _{x \to \infty } \frac {\ln (x)}{1 + x^2} = \answer {0}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\ln (x)$ is $\frac {1}{x}$

Simplify the new fraction and then compute the limit

The $0 \cdot \infty$ case

L’Hopital’s Rule requires a fractional indeterminate form such as $\frac 00$ or $\frac {\infty }{\infty }$ , but we can use it to handle other indeterminate forms by rewriting expressions as fractions.

Examples of the $0\cdot \infty$ case.

example 8 Compute the limit: $\displaystyle {\;\lim _{x \to 0^+}x^2 \ln (x)}$ .

As $x\to 0^+$ we get $0 \cdot (-\infty )$ which is an indeterminate form, but L’Hopital’s Rule does not apply in this situation. We must rewrite the problem as a fraction, in the following way: $\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}.$ Notice that this is equivalent to the original problem since $x^2 = \frac {1}{x^{-2}}.$ Also note that $x^{-2} =\frac {1}{x^2} \to \infty$ as $x\to 0^+$ . Now, we can use L’Hopital’s Rule because $\lim _{x \to 0+}\frac {\ln (x)}{x^{-2}}= \frac {-\infty }{\infty }.$ We get $\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}= \lim _{x\to 0^+} \frac {1/x}{-2x^{-3}}$ which simplifies to $\lim _{x \to 0^+}-\frac {x^2}{2}= 0.$ Hence, $\lim _{x \to 0^+} x^2 \ln (x)=\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}= 0.$

(problem 8a) Compute $\lim _{x \to 0^+} x^3\ln (x) = \answer {0}$

(problem 8b) Compute $\lim _{x \to \infty } 3xe^{-x} = \answer {0}$

“Plug in” $x=\infty$

If you got $\infty \cdot 0$ , rewrite the expression as a fraction

Take advantage of negative exponents: $e^{-x} = \frac {1}{e^x}$

Here is a detailed, lecture style video on L’Hopital’s Rule:

2024-09-27 13:56:19

1 L’Hopital’s Rule

2 The indeterminate form \frac 00

3 The \dfrac {\infty }{\infty } case.

4 The 0 \cdot \infty case

The indeterminate form $\frac 00$

The $\dfrac {\infty }{\infty }$ case.

The $0 \cdot \infty$ case