6.11 L’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).

L’Hôpital’s Rule for the Indeterminate Form 0/0

Assume \(f\) and \(g\) are two functions with \(f(a)=g(a)=0\). Then for \(x\neq a\), we have \[\begin{aligned} \frac{f(x)}{g(x)} & =\frac{f(x)-f(a)}{g(x)-g(a)}\\ \\ & =\frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}\end{aligned}\] Suppose the derivatives \(f^\prime(a)\) and \(g^\prime(a)\) exist and \(g^\prime(a)\neq0\). Because \[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=f^\prime(a),\qquad(h=x-a)\] and \[\lim_{x\to a}\frac{g(x)-g(a)}{x-a}=g^\prime(a),\] we get \[\boxed{\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^\prime(a)}{g^\prime(a)}\tag{i}}\] provided that \(g^\prime(a)\neq0\).

Figure 1 visually justifies Equation (i). Figure 1(a) shows the graphs of two differentiable functions \(f\) and \(g\) near \(x=a\). If we zoom in toward the point \((a,0)\) (Figure 1(b)), the graphs of \(f\) and \(g\) look almost linear, so we can approximate the ratio of \(f\) and \(g\) by the ratio of their local linearizations. That is, because \[f(x)\approx L_{f}(x)=\underbrace{f(a)}_{=0}+f^\prime(a)(x-a)\] and \[g(x)\approx L_{g}(x)=\underbrace{g(a)}_{=0}+g^\prime(a)(x-a),\] we have \[\frac{f(x)}{g(x)}\approx\frac{f^\prime(a)(x-a)}{g^\prime(a)(x-a)}=\frac{f^\prime(a)}{g^\prime(a)}.\]

(a)	(b)

Figure 1

Use Equation (i) only when \(f(a)=g(a)=0\). If this condition is not satisfied, we cannot use this equation.

For example, consider \[\lim_{x\to0}\frac{1+2x}{1-x}\] Here \(f(x)=1+2x\) and \(g(x)=1-x\). Because \(f(0)/g(0)\neq0/0\), we cannot use Equation (i). If we use (i), because \(f^\prime(0)=2\) and \(g^\prime(0)=-1\), we get \(2/(-1)=-2\). But the correct answer can be obtained simply by subsituting 0 for \(x\): \[\lim_{x\to0}\frac{1+2x}{1-x}=\frac{1+2\times0}{1-0}=1\neq\frac{f^\prime(0)}{g^\prime(0)}=\frac{2}{-1}=-2.\]

• 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 \(-\infty\).

By the hypothesis that \({\displaystyle \lim_{x\to a}}f^\prime(x)/g^\prime(x)\) exists, it is tactically assumed there is an open interval \(I\) containing \(a\) such that

1. Both \(f\) and \(g\) are differentiable (\(f^\prime(x)\) and \(g^\prime(x)\) exist) for all \(x\) in \(I\) (except possibly for \(x=a\)).
2. \(g^\prime(x)\neq0\) in \(I\) (except possibly when \(x=a\)).

Show the proof …

Hide the proof

• Notice that \(f^\prime(x)/g^\prime(x)\) is the derivative of the numerator divided by the derivative of the denominator and it is different from the derivative of the fraction \(f(x)/g(x)\). \[\frac{f^\prime(x)}{g^\prime(x)}\neq\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f^\prime(x)g(x)-g^\prime(x)f(x)}{[g(x)]^{2}}.\]
• If we apply l’Hôpital’s rule to the well-known limit \(\lim_{x\to0}\sin x/x\), we obtain \[\lim_{x\to0}\frac{\sin x}{x}\stackrel{H}{=}\lim_{x\to0}\frac{\cos x}{1}=\cos0=1\] Although we could obtain this limit easily, to derive the formula \(\frac{d}{dx}\sin x=\cos x\), we assumed the truth of this limit.
• In these notes, we place an “H” above the equal sign to indicate that the two limits are equal as a result of applying l’Hôpital’s rule.

  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 \(x\).

Again note that if \(f,g\to0\) as \(x\to s\), but \[\lim_{x\to s}\frac{f^\prime(x)}{g^\prime(x)}\] does not approach a limit (that is, it is not a finite number, \(+\infty\), or \(-\infty\)), we cannot conclude that \(\lim_{x\to s}f(x)/g(x)\) does not approach a limit. The next example shows such a situation.

L’Hópital’s Rule for the Indeterminate Form  ±∞/±∞

The proof of this theorem is discussed in more advanced books.

• The above example shows us that \(x^{\alpha}\) (\(\alpha>0\)) grows much faster than \(\ln x\).

Solution 16

Because \[\lim_{x\to0^{+}}\ln x=-\infty\]

 

and \[\lim_{x\to0^{+}}\csc x=\lim_{x\to0^{+}}\frac{1}{\sin x}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}+\infty\] we have an indeterminate limit of type \(-\infty/\infty\) and we can apply l’Hôpital’s rule: \[\lim_{x\to0^{+}}\frac{\ln x}{\csc x}\stackrel{H}{=}\lim_{x\to0^{+}}\frac{\dfrac{1}{x}}{-\csc x\cot x}\] The second limit is still indeterminate of type \(\infty/(-\infty)\). If we directly apply l’Hôpital’s rule to this fraction, because powers of \(1/x\) appear in the numerator and expressions involving \(\csc x\) and \(\cot x\) in the denominator, we will make the problem more complicated. However, if we first simplify the quotient, l’Hôpital’s rule is then helpful: \[\begin{aligned} \lim_{x\to0^{+}}\frac{\dfrac{1}{x}}{-\csc x\cot x} & =\lim_{x\to0^{+}}\frac{-1}{x\dfrac{1}{\sin x}\dfrac{\cos x}{\sin x}}\\ \\ & \begin{equation*}=\lim_{x\to0^{+}}\frac{\sin^{2}x}{x\cos x}\tag{0/0}\end{equation*}\\ \\ & \stackrel{H}{=}\lim_{x\to0^{+}}\frac{2\sin x\cos x}{\cos x-x\sin x}\\ & =\frac{2(0)(1)}{1-0(0)}=0.\end{aligned}\]

Evaluation of the Indeterminate Form 0·(±∞)

We can apply l’Hôpital’s rule to evaluate the limits of the indeterminate form \(0\cdot(\pm\infty)\). Namely,

Solution 17

Because \[\begin{aligned} \lim_{x\to\pi/2}\sec3x & =\lim_{x\to\pi/2}\frac{1}{\cos3x}\\ & =\begin{cases} \lim_{x\to\frac{\pi}{2}^{-}}\frac{1}{\cos3x}\stackrel{\left[\frac{1}{0^{-}}\right]}{=}-\infty & \text{(see Figure 3)}\\ \lim_{x\to\frac{\pi}{2}^{+}}\frac{1}{\cos3x}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}+\infty & \text{(see Figure 3)} \end{cases}\end{aligned}\] and \[\lim_{x\to\pi/2}\cos5x=\cos(5\pi/2)=0.\]

 

 

That is, we have an indeterminate limit of type \(0\cdot\infty\) or \(0\cdot(-\infty)\) (Figure 4). If we write, \[\sec3x\cos5x=\frac{\cos5x}{\cos3x},\] we get an indeterminate limit of type \(0/0\): \[\lim_{x\to\pi/2}\frac{\cos5x}{\cos3x}=\left[\frac{\cos(5\pi/2)}{\cos(3\pi/2)}=\frac{0}{0}\right]\]

 

Now we can apply l’Hôpital’s rule. \[\frac{d}{dx}\cos5x=-5\sin5x\] \[\frac{d}{dx}\cos3x=-3\sin3x\] \[\begin{aligned} \Rightarrow\lim_{x\to\pi/2}\frac{\cos5x}{\cos3x} & \overset{H}{=}\lim_{x\to\pi/2}\frac{-5\sin5x}{-3\sin3x}\\ \\ & =\frac{-5\overbrace{\sin(5\pi/2)}^{=\sin(2\pi+\frac{\pi}{2})=1}}{-3\underbrace{\sin(3\pi/2)}_{=-1}}\\ & =-\frac{5}{3}.\end{aligned}\] Therefore, \(\sec3x\cos5x\to-5/3\) as \(x\to\pi/2\). The graph of \(y=\sec3x\cos5x\) is shown in Figure 5.

Solution 18

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 \(0\cdot(\pm\infty)\) (Figure 6).

 

We can transform this limit into one of the form \(0/0\), and then apply l’Hôpital’s rule: \[\begin{aligned} \lim_{x\to\frac{\pi}{2}}(\pi-2x)\tan x & =\lim_{x\to\frac{\pi}{2}}\frac{\pi-2x}{\cot x}\\ & \stackrel[H]{\left[\frac{0}{0}\right]}{=}\lim_{x\to\frac{\pi}{2}}\frac{-2}{-(1+\cot^{2}x)}\\ \\ & =\frac{-2}{-(1+\cot^{2}(\pi/2))}\\ \\ & =\frac{2}{1+0}=2.\end{aligned}\] The graph of \(y=(\pi-2x)\tan x\) is shown in Figure 7.

Solution 19

Note that \[\sin x\to0^{+}\text{ as }x\to0^{+},\] and \[\ln u\to-\infty\text{ as }u\to0^{+}.\] Therefore, \[\lim_{x\to0^{+}}\ln\sin x=-\infty\] and \(\lim_{x\to0^{+}}x\ln\sin x\) is a limit of the form \(0\cdot(-\infty)\) (see Figure 8).

 

We can transform it into one of the form \(\infty/\infty\). \[\begin{aligned} \lim_{x\to0^{+}}x\ln x & =\lim_{x\to0^{+}}\frac{\ln\sin x}{\frac{1}{x}}\left[=\frac{-\infty}{+\infty}\right]\\ & \overset{H}{=}\lim_{x\to0^{+}}\frac{\cos x\ \dfrac{1}{\sin x}}{-\dfrac{1}{x^{2}}}\\ & =\lim_{x\to0^{+}}x^{2}\cos x\ \frac{1}{\sin x}\\ & =\underbrace{\left(\lim_{x\to0^{+}}\cos x\right)}_{=1}\underbrace{\left(\lim_{x\to0^{+}}x\right)}_{=0}\underbrace{\left(\lim_{x\to0^{+}}\frac{x}{\sin x}\right)}_{=1}\\ & =1(0)(1)=0.\end{aligned}\] [Recall that \(\sin x/x\to0\) as \(x\to0\).] The graph of \(y=x\ln\sin x\) is shown in Figure 9.

Solution 20

Because \(e^{x}\to0\) as \(x\to-\infty\), we have an indeterminate limit of type \(0\cdot(-\infty)\). We write it as \[\begin{aligned} \lim_{x\to-\infty}xe^{x} & =\lim_{x\to-\infty}\frac{x}{e^{-x}}\left[=\frac{-\infty}{+\infty}\right]\\ & \overset{H}{=}\lim_{x\to-\infty}\frac{1}{-e^{-x}}\\ & =-\lim_{x\to-\infty}e^{x}\\ & =0.\end{aligned}\] Figure 10 shows the graph of \(y=xe^{x}\).

Evaluation of the Indeterminate Form ∞-∞

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 difference into a fraction are:

• Using a common denominator
• Rationalization
• Factoring out a common factor

or we may write:

\[\begin{aligned} f(x)-g(x) & =\frac{1}{\dfrac{1}{f(x)}}-\frac{1}{\dfrac{1}{g(x)}}\\ \\ & =\frac{\dfrac{1}{g(x)}-\dfrac{1}{f(x)}}{\dfrac{1}{f(x)g(x)}}=\frac{0}{0}\end{aligned}\]

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 \(g\to0\).

In any of the above cases, \(\lim_{x\to s}(g(x)\ln f(x))\) will take on the indeterminate form \(0\cdot\pm\infty\).

• Note that \(0^{\pm\infty}\) is not an indeterminate form. If \(f\to0^{+}\) and \(g\to+\infty\) , then \[\ln f(x)\to-\infty\] and \[g(x)\ln f(x)\to(+\infty)(-\infty)=-\infty\] and \[y=e^{g(x)\ln f(x)}\stackrel{e^{-\infty}}{\to}0.\] And if \(f\to0^{+}\) and \(g\to-\infty\), then \[g(x)\ln f(x)\to(-\infty)(-\infty)=+\infty\] and \[y=e^{g(x)\ln f(x)}\stackrel{e^{+\infty}}{\to}+\infty.\]

Solution 24

Let \[y=x^{x};\] then \[\ln y=x\ln x.\] As \(x\to0^{+}\), \(x\ln x\) takes on the form \(0\cdot(-\infty)\). To find \(\lim_{x\to0^{+}}x\ln x\), we write \[\begin{aligned} \lim_{x\to0^{+}}x\ln x & =\lim_{x\to0^{+}}\frac{\ln x}{\frac{1}{x}}\left[=\frac{-\infty}{+\infty}\right]\\ & \stackrel{H}{=}\lim_{x\to0^{+}}\frac{\frac{1}{x}}{-\frac{1}{x^{2}}}\\ & =\lim_{x\to0^{+}}(-x)=0.\end{aligned}\] Therefore, \[\ln y\to0\text{ as }x\to0^{+}.\] So far we have obtained the limit of \(\ln y\). To compute the limit of \(y\), we use the fact that \(y=e^{\ln y}\) and \[\lim_{x\to0^{+}}y=\lim_{x\to0^{+}}e^{\ln y}\] and the continuity of the exponential function implies \[\lim_{x\to0^{+}}e^{\ln y}=e^{\lim_{x\to0^{+}}\ln y}=e^{0}=1\] so \[y\to1\text{ as }x\to0^{+}.\]

Solution 26

This function assumes the indeterminate form \(1^{\pm\infty}\). Let \[y=(1+\sin x)^{1/x}\] \[\Rightarrow\ln y=\frac{1}{x}\ln(1+\sin x)\quad(\ln a^{b}=b\ln a)\] and \[\begin{aligned} \lim_{x\to0}\ln y & =\lim_{x\to0}\frac{\ln(1+\sin x)}{x}\left[=\frac{0}{0}\right]\\ & \stackrel{H}{=}\lim_{x\to0}\frac{\cos x\ \frac{1}{1+\sin x}}{1}\\ & =\cos0\left(\frac{1}{1+\sin0}\right)\\ & =1\end{aligned}\] Because \(y=e^{\ln y}\) and \(\ln y\to1\) as \(x\to0\), we have \[y\to e^{1}=e\text{ as }x\to0.\]