3.6 L’Hopital’s Rule

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

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

If $\displaystyle {\lim _{x \to c} \frac {f(x)}{g(x)} = \frac {0}{0} \quad \text {or} \quad \frac {\infty }{\infty }}$
then $\displaystyle {\lim _{x \to c} \frac {f(x)}{g(x)} = \lim _{x \to c} \frac {f'(x)}{g'(x)}}$ provided the latter exists.
In the above statement, $\lim _{x \to c}$ can be replaced by a one-sided limit and $c$ can be $\pm \infty$ . Also the fraction $\frac {\infty }{\infty }$ is shorthand for $\frac {\pm \infty }{\pm \infty }$ .

We begin with the $\frac 00$ form.

Compute the limit: $\displaystyle {\lim _{x \to 0} \frac {\sin (x)}{x}}.$

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

We have

$\lim _{x \to 0} \frac {\sin (x)}{x} = \lim _{x \to 0} \frac {\cos (x)}{1} = 1.$

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

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

Plug in $x=0$

If you got $0/0$ , 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

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

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

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

Plug in $x=0$

If you got $0/0$ , 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

Compute the limit: $\displaystyle {\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32}}.$
Plugging in the terminal value, $x=2$ , yields the indeterminate form $0/0$ , so L’Hopital’s rule applies. We have $\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32} = \lim _{x \to 2} \frac {3x^2}{5x^4} = \frac {12}{80} = \frac {3}{20}.$

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

Plug in $x=3$

If you got $0/0$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

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

Plug in $x=4$

If you got $0/0$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

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

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 $0/0$ , 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 $\displaystyle {\lim _{x \to 0} \frac {e^x - x - 1}{x^2} = \frac 12}.$

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

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

Plug in $x=0$

If you got $0/0$ , 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 $\sec (x)\tan (x)$

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 $\infty /\infty$ . These are handled the same way as the $0/0$ case above.

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

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 $x \to \infty$ .

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

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

Compute the limit: $\displaystyle {\lim _{x \to \infty } \frac {\sqrt {x}}{\ln (x)}}$ . As $x \to \infty$ we get $\infty /\infty$ , 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 $x\to \infty$ .

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

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

Other Indeterminate Forms

L’Hopital’s Rule requires a fractional indeterminate form such as $0/0$ or $\infty /\infty$ , but we can use it to handle other indeterminate forms by rewriting expressions as fractions.

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

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, $\lim _{x \to 0^+} x^2 \ln (x)=\lim _{x \to 0^+}\frac {ln(x)}{x^{-2}}= 0.$

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: