Short and Sweet Calculus

## 4.6L’Hôpital’s Rule for Indeterminate Limits

In this section, we learn a powerful method to attack problems on indeterminate forms, called l’Hôpital’s rule (also written l’Hospital’s rule).

### 4.6.1 L’Hôpital’s Rule for the Indeterminate Forms of Type $$0/0$$ and $$\pm\infty/\pm\infty$$

Assume $$f$$ and $$g$$ are two functions with $$f(a)=g(a)=0$$. Then for $$x\neq a$$, we have $\frac{f(x)}{g(x)}=\frac{f(x)-\overbrace{f(a)}^{0}}{g(x)-\underbrace{g(a)}_{0}}=\frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}$ Suppose the derivatives $$f'(a)$$ and $$g'(a)$$ exist and $$g'(a)\neq0$$. Because $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=f'(a),\qquad(h=x-a)$ and $\lim_{x\to a}\frac{g(x)-g(a)}{x-a}=g'(a),$ we get $\boxed{\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}\tag{i}}$ provided that $$g'(a)\neq0$$.

The following theorem generalizes the cases where we can use Equation (i). It states that Equation (i) is true under less stringent conditions and is also valid for one-sided limits as well as limits at $$+\infty$$ and $$-\infty$$.

4.7. (l’Hôpital’s Rule): If $$f$$ and $$g$$ both approach $$0$$ or both approach $$\pm\infty$$ as $$x\to s$$, then $\lim_{x\to s}\frac{f(x)}{g(x)}=\lim_{x\to s}\frac{f'(x)}{g'(x)}$ provided that $${\displaystyle \lim_{x\to s}}\frac{f'(x)}{g'(x)}$$ exists or the limit is $$+\infty$$ or $$-\infty$$.

Here $$s$$ signifies $$a,a^{+},a^{-},-\infty$$, or $$+\infty$$, where $$a$$ is a real number.

To apply l’Hôpital’s rule:

1. Make sure the limit of $$f(x)/g(x)$$ assumes the form $$\dfrac{0}{0}$$ or $$\dfrac{\pm\infty}{\pm\infty}$$ as $$x\to s$$.

2. Differentiate $$f(x)$$ and $$g(x)$$ separately.

3. Find the limit of $$f'(x)/g'(x)$$ as $$x\to s$$. If the limit is a number, $$+\infty$$, or $$-\infty$$, then it is equal to the limit of $$f(x)/g(x)$$; otherwise, we CANNOT conclude that the limit of $$f(x)/g(x)$$ does not exist.

• If necessary, we may repeat the above process. Stop differentiating as soon as the derivative of the numerator or that of the denominator is different from zero or infinity.

Example 4.7. Find $\lim_{x\to1}\frac{x^{3}-3x+2}{2x^{3}-x^{2}-4x+3}.$

Solution

Because $\left.\frac{x^{3}-3x+2}{2x^{3}-x^{2}-4x+3}\right|_{x=1}=\frac{1-3+2}{2-1-4+3}=\frac{0}{0},$ we can apply l’Hôpital’s rule $\lim_{x\to1}\frac{x^{3}-3x+2}{2x^{3}-x^{2}-4x+3}\stackrel{H}{=}\lim_{x\to1}\frac{3x^{2}-3}{6x^{2}-2x-4}.$ Because $\left.\frac{3x^{2}-3}{6x^{2}-2x-4}\right|_{x=1}=\frac{3(1)^{2}-3}{6(1)^{2}-2(1)-4}=\frac{0}{0},$ we apply l’Hôpital’s rule again $\lim_{x\to1}\frac{3x^{2}-3}{6x^{2}-2x-4}\stackrel{H}{=}\lim_{x\to1}\frac{6x}{12x-2}=\frac{6(1)}{12(1)-2}=\frac{3}{5}.$

• It is a common error to apply l’Hôpital’s rule to calculate the limit of $$6x/(12x-2)$$ as $$x\to1$$. Applying l’Hôpital’s rule leads to a wrong value because this limit is not indeterminate and can be simply obtained by substituting 1 for $$x$$.

Example 4.8. Find $\lim_{x\to1}\frac{\ln x}{x-1}.$

Solution

Because $\left.\frac{\ln x}{x-1}\right|_{x=1}=\frac{0}{0},$ we can apply l’Hôpital’s rule: $\lim_{x\to1}\frac{\ln x}{x-1}\stackrel{H}{=}\lim_{x\to1}\frac{\dfrac{1}{x}}{1}=1.$

Example 4.9. Find $\lim_{x\to\frac{\pi}{2}^{+}}\frac{\tan5x}{\tan x}.$

Solution

Because $\lim_{x\to\frac{\pi}{2}^{+}}\tan5x=\lim_{x\to\frac{\pi}{2}^{+}}\tan x=-\infty,$ we have \begin{aligned} \lim_{x\to\frac{\pi}{2}^{+}}\frac{\tan5x}{\tan x} & \stackrel{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{5\sec^{2}5x}{\sec^{2}x}\\ & =\lim_{x\to\frac{\pi}{2}^{+}}\frac{5\cos^{2}x}{\cos^{2}5x}\tag{\small0/0}\\ & \overset{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{\cancel{5(2)}(\bcancel{-}\sin x)\cos x}{\cancel{2(5)}(\bcancel{-}\sin5x)\cos5x}\\ & =\left(\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\sin5x}\right)\left(\lim_{x\to\frac{\pi}{2}^{+}}\frac{\cos x}{\cos5x}\right).\end{aligned} Because $$\sin\frac{\pi}{2}=1$$ and $$\sin\frac{5\pi}{2}=\sin\left(2\pi+\frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1$$, we have: $\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\sin5x}=1.$ Because $$\cos\frac{\pi}{2}=0$$ and $$\cos\frac{5\pi}{2}=\cos\left(2\pi+\frac{\pi}{2}\right)=\cos\frac{\pi}{2}=0$$, we have an indeterminate limit of type $$0/0$$. So we can apply l’Hôpital’s rule: $\lim_{x\to\frac{\pi}{2}^{+}}\frac{\cos x}{\cos5x}\stackrel{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{-\sin x}{-5\sin5x}=\frac{1}{5}.$ Therefore $\lim_{x\to\frac{\pi}{2}^{+}}\frac{\tan5x}{\tan x}=\left(\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\sin5x}\right)\left(\lim_{x\to\frac{\pi}{2}^{+}}\frac{\cos x}{\cos5x}\right)=(1)\left(\frac{1}{5}\right)=\frac{1}{5}.$

Example 4.10. Evaluate $\lim_{x\to-\infty}\frac{\sin(1/x)}{\ln\frac{x+1}{x+2}}.$

Solution

Because $$\sin x$$ and $$\ln x$$ are continuous functions, we have $\lim_{x\to-\infty}\sin\frac{1}{x}=\sin\left(\lim_{x\to-\infty}\frac{1}{x}\right)=\sin0=0$ and $\lim_{x\to-\infty}\ln\frac{x+1}{x+2}=\ln\left(\lim_{x\to-\infty}\frac{x+1}{x+2}\right)=\ln1=0.$ So we have an indeterminate limit of type 0/0, and l’Hôpital’s rule applies: \begin{aligned} \lim_{x\to-\infty}\frac{\sin(1/x)}{\ln\frac{x+1}{x+2}} & \overset{H}{=}\lim_{x\to-\infty}\frac{-\frac{1}{x^{2}}\cos\frac{1}{x}}{\underbrace{\frac{1(x+2)-1(x+1)}{(x+2)^{2}}}_{u’}\underbrace{\frac{x+2}{x+1}}_{1/u}}\tag{\ensuremath{u=\frac{x+1}{x+2}}}\\ & =\lim_{x\to-\infty}\frac{-\frac{1}{x^{2}}\cos\frac{1}{x}}{\frac{1}{(x+2)}\frac{1}{(x+1)}}\\ & =\lim_{x\to-\infty}\left(-\frac{(x+1)(x+2)}{x^{2}}\cos\frac{1}{x}\right)\\ & =-\left(\lim_{x\to-\infty}\frac{(x+1)(x+2)}{x^{2}}\right)\left(\lim_{x\to-\infty}\cos\frac{1}{x}\right)\end{aligned} The limit of a rational function as $$x\to+\infty$$ or $$-\infty$$ is the same as the limit of ration of the highest degree term in the numerator and the highest degree term in the denominator. So $\lim_{x\to-\infty}\frac{x^{2}+3x+2}{x^{2}}=\lim_{x\to-\infty}\frac{x^{2}}{x^{2}}=1.$ Also $\lim_{x\to-\infty}\cos\frac{1}{x}=\cos\left(\lim_{x\to-\infty}\frac{1}{x}\right)=\cos0=1.$ Therefore $\lim_{x\to-\infty}\frac{\sin(1/x)}{\ln\frac{x+1}{x+2}}=-\left(\lim_{x\to-\infty}\frac{(x+1)(x+2)}{x^{2}}\right)\left(\lim_{x\to-\infty}\cos\frac{1}{x}\right)=-(1)(1)=-1.$

### 4.6.2 Evaluation of the Indeterminate Form $$0\cdot(\pm\infty)$$

If $$f(x)\to0$$ and $$g(x)\to+\infty$$ or $$-\infty$$ as $$x\to s$$, then we write $f(x)g(x)=\frac{f(x)}{1/g(x)}\left(\text{or }=\frac{g(x)}{1/f(x)}\right)$ so as to cause it to take on one of the forms $$0/0$$ or $$\pm\infty/\pm\infty$$, and then apply l’Hôpital’s rule.

Example 4.11. Evaluate $\lim_{x\to\pi/2}(\pi-2x)\tan x.$

Solution

Note that because $\lim_{x\to\pi/2^{+}}\tan x=-\infty,\quad\lim_{x\to\pi/2^{-}}\tan x=+\infty$ we have an indeterminate limit of type $$0\cdot(\pm\infty)$$.

We can transform this limit into one of the form $$0/0$$, $\lim_{x\to\frac{\pi}{2}}(\pi-2x)\tan x=\lim_{x\to\frac{\pi}{2}}\frac{\pi-2x}{\cot x}\quad\left[=\frac{0}{0}\right]$ and then apply l’Hôpital’s rule: $\lim_{x\to\frac{\pi}{2}}\frac{\pi-2x}{\cot x}\stackrel{H}{=}\lim_{x\to\frac{\pi}{2}}\frac{-2}{-(1+\cot^{2}x)}=\frac{-2}{-(1+\cot^{2}(\pi/2))}=\frac{2}{1+0}=2.$

Example 4.12. Evaluate $\lim_{x\to-\infty}xe^{x}.$

Solution

Because $$e^{x}\to0$$ as $$x\to-\infty$$, we have an indeterminate limit of type $$0\cdot(-\infty)$$. We write it as \begin{aligned} \lim_{x\to-\infty}xe^{x} & =\lim_{x\to-\infty}\frac{x}{e^{-x}}\left[=\frac{-\infty}{+\infty}\right]\\ & \overset{H}{=}\lim_{x\to-\infty}\frac{1}{-e^{-x}}\\ & =-\lim_{x\to-\infty}e^{x}\\ & =0.\end{aligned}

### 4.6.3 Evaluation of the indeterminate form $$\infty-\infty$$

To evaluate a limit of the indeterminate form $$\infty-\infty$$, we can transform it into a limit of a fraction such that we get an indeterminate form of type $$0/0$$ or $$\pm\infty/\pm\infty$$, then we apply l’Hôpital’s rule. Some examples of converting a a difference into a fraction are:

• Using a common denominator

• Rationalization

• Factoring out a common factor

Example 4.13. Evaluate $\lim_{x\to0}\left(\frac{1}{\sin x}-\frac{1}{x}\right).$

Solution

Here we have a limit of the form $$\infty-\infty$$ (you may consider $$x\to0^{+}$$ and $$x\to0^{-}$$ separately). Using a common denominator, we get $\lim_{x\to0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\lim_{x\to0}\frac{x-\sin x}{x\sin x}\left[=\frac{0}{0}\right]$ Now we can apply l’Hôpital’s rule: \begin{aligned} \lim_{x\to0}\frac{x-\sin x}{x\sin x} & \overset{H}{=}\lim_{x\to0}\frac{1-\cos x}{\sin x+x\cos x}\left[=\frac{0}{0}\right]\\ & \overset{H}{=}\lim_{x\to0}\frac{+\sin x}{\cos x+\cos x-x\sin x}\\ & =\frac{\sin0}{2\cos0-0(0)}\\ & =0.\end{aligned}

### 4.6.4 Evaluation of the indeterminate forms $$0^{0},1^{\pm\infty},(\pm\infty)^{0}$$

In addition to $$0/0$$, $$\pm\infty/\infty$$, $$0\cdot(\pm\infty)$$, and $$\infty-\infty$$, other indeterminate forms are $0^{0},1^{\pm\infty},\text{ and }(\pm\infty)^{0}.$ That is, limits of the form

$\lim_{x\to s}f(x)^{g(x)}\quad[\text{with }f(x)>0]$ are indeterminate if

• $$f\to0$$ and $$g\to0$$

• $$f\to1$$ and $$g\to+\infty$$ or $$-\infty$$

• $$f\to+\infty$$ or $$-\infty$$ and $$g\to0$$.

To evaluate such limits, let

$y=f(x)^{g(x)}.$ Taking the natural logarithm of each side, \begin{aligned} \ln y & =\ln f(x)^{g(x)}\\ & =g(x)\ln f(x),\tag{Recall \ensuremath{\ln(a^{b})=b\ln a}}\end{aligned} and then raising $$e$$ to the power of both sides, we obtain $\underbrace{e^{\ln y}}_{=y}=e^{g(x)\ln f(x)}.$ Because $$e^{x}$$ is a continuous function $\lim_{x\to s}y=e^{\lim_{x\to s}\left(g(x)\ln f(x)\right)}.$ Therefore: $\boxed{\lim_{x\to s}f(x)^{g(x)}=e^{\lim_{x\to s}\left(g(x)\ln f(x)\right)}.}$ In any of the above cases, $$\lim_{x\to s}(g(x)\ln f(x))$$ will take on the indeterminate form $$0\cdot\pm\infty$$.

Example 4.14. Evaluate $\lim_{x\to0^{+}}x^{x}.$

Solution

Let $$y=x^{x}$$. Taking the natural logarithn of both sides, we get $\ln y=x\ln x.$ Therefore, $y=e^{x\ln x}$ As $$x\to0^{+}$$, $$x\ln x$$ takes on the form $$0\cdot(-\infty)$$. To find $${\displaystyle \lim_{x\to0^{+}}x\ln x}$$, we write it as $\lim_{x\to0^{+}}x\ln x=\lim_{x\to0^{+}}\frac{\ln x}{\dfrac{1}{x}}\quad\left[=\frac{-\infty}{+\infty}\right]$ Now we can apply l’Hôpital’s rule: $\lim_{x\to0^{+}}\frac{\ln x}{\dfrac{1}{x}}\stackrel{H}{=}\lim_{x\to0^{+}}\frac{\frac{1}{x}}{-\frac{1}{x^{2}}}=\lim_{x\to0^{+}}(-x)=0.$ Therefore $\lim_{x\to0^{+}}x^{x}=\lim_{x\to0^{+}}y=e^{\lim_{x\to0^{+}}(x\ln x)}=e^{0}=1.$