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 justify 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 \[\frac{f(x)}{g(x)}\approx\frac{f^\prime(a)(x-a)}{g^\prime(a)(x-a)}=\frac{f^\prime(a)}{g^\prime(a)}.\]
Here \[f(x)=\sin x,\qquad g(x)=1-e^{x},\] with \[f(0)=0,\qquad g(0)=0.\] Because \[f^\prime(x)=\cos x\Rightarrow f^\prime(0)=\cos0=1\] and \[g^\prime(x)=-e^{x}\Rightarrow g^\prime(0)=-e^{0}=-1.\] by Equation (i), we have \[\lim_{x\to0}\frac{\sin x}{1-e^{x}}=\frac{f^\prime(0)}{g^\prime(0)}=\frac{1}{-1}=-1.\]
Here \(f(x)=1-x^{3}\) and \(g(x)=1-x\) with \(f(1)=0\) and \(g(0)=0\). We can use Equation (i) to evaluate the above limit: \[f(x)=1-x^{3}\Rightarrow f^\prime(x)=-3x^{2}\Rightarrow f^\prime(1)=-3\]\[g(x)=1-x\Rightarrow g^\prime(x)=-1\Rightarrow g^\prime(1)=-1\]\[\lim_{x\to1}\frac{1-x^{3}}{1-x}=\frac{f^\prime(1)}{g^\prime(1)}=\frac{-3}{-1}=3.\]
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 obstained 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 \(-\infty\).
Theorem 1.(l’Hôpital’s Rule): Assume \[\lim_{x\to s}f(x)=0\quad\text{and}\quad\lim_{x\to s}g(x)=0\] and assume also that \({\displaystyle \lim_{x\to s}}f^\prime(x)/g^\prime(x)\) exists or this limit is \(+\infty\) or \(-\infty\). Then \({\displaystyle \lim_{x\to s}}f(x)/g(x)\) also approaches a limit and \[\lim_{x\to s}\frac{f(x)}{g(x)}=\lim_{x\to s}\frac{f^\prime(x)}{g^\prime(x)}.\] Here \(s\) signifies \(a,a^{+},a^{-},-\infty\), or \(+\infty\), where \(a\in\mathbb{R}\).
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
Both \(f\) and \(g\) are differentiable (\(f^\prime(x)\) and \(g^\prime(x)\) exist) for all \(x\) in \(I\) (except possibly for \(x=a\)).
\(g^\prime(x)\neq0\) in \(I\) (except possibly when \(x=a\)).
Show the proof …
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First we prove this theorem for \(x\to a^{+}\). Because \(f\) and \(g\) may not be defined at \(a\), we introduce two new functions \[F(x)=\begin{cases} f(x) & \text{if }x\neq a\\ 0 & \text{if }x=a \end{cases}\quad G(x)=\begin{cases} g(x) & \text{if }x\neq a\\ 0 & \text{if }x=a \end{cases}.\] Both \(F\) and \(G\) are continuous at \(a\) because \[\lim_{x\to a}F(x)=\lim_{x\to a}f(x)=0=F(a)\] and similarly \(\lim_{x\to a}G(x)=0=G(0)\).
If \(a<x\) and \(x\in I\), then \(F\) and \(G\) are continuous on \([a,x]\) and differentiable on \((a,x)\) (because \(f^\prime=f^\prime\) and \(g^\prime=g^\prime\)). Therefore, we can apply Cauchy’s Mean-Value Formula to the interval \([a,x]\) and obtain \[\left[\underbrace{F(x)}_{=f(x)}-\overset{=0}{\cancel{F(a)}}\right]\underbrace{g^\prime(c)}_{=g^\prime(c)}=\left[\underbrace{G(x)}_{=g(x)}-\overset{=0}{\cancel{G(a)}}\right]\underbrace{f^\prime(c)}_{=f^\prime(c)}\] or \[f(x)g^\prime(c)=g(x)f^\prime(c)\tag{*}\] for some \(c\) in the interval \((a,x)\).
Because \(g^\prime(x)\neq0\) for all \(x\) in \(I-\{a\}\), we have:
(1) \(g^\prime(c)\neq0\) (\(a<c<x\))
(2) \(g(x)\neq0\), because if \(g(x)=0\), then we would have \(g(x)=g(a)=0\) and by Rolle’s theore, there would be a point \(c_{1}\) between \(a\) and \(x\) such that \(g^\prime(c_{1})=0\).
Therefore, we divide both sides of (*) by \(g^\prime(c)\) and \(g(x)\) and obtain \[\frac{f(x)}{g(x)}=\frac{f^\prime(c)}{g^\prime(c)}.\] Now if we let \(x\to a^{+}\), then \(c\to a^{+}\) (because \(a<c<x\)), so \[\lim_{x\to a^{+}}\frac{f(x)}{g(x)}=\lim_{c\to a^{+}}\frac{f^\prime(c)}{g^\prime(c)}.\] The method needs some minor modifications to show that the result is valid as \(x\to a^{-}\). The combination of these two one-sided limit cases proves that the theorem is true as \(x\to a\).
When \(x\to+\infty\), the substitution \(x=\frac{1}{u}\) reduces the limit to evaluation of the limit as \(u\to0^{+}\). Thus \[\lim_{x\to+\infty}\frac{f(x)}{g(x)}=\lim_{u\to0^{+}}\frac{f(1/u)}{g(1/u)}\] Applying l’Hôpital’s rule: \[\begin{aligned} \lim_{u\to0^{+}}\frac{f(1/u)}{g(1/u)} & =\lim_{u\to0^{+}}\frac{-\dfrac{1}{u^{2}}f^\prime\left(\dfrac{1}{u}\right)}{-\dfrac{1}{u^{2}}g^\prime\left(\dfrac{1}{u}\right)}\\ & =\lim_{u\to0^{+}}\frac{f^\prime(1/u)}{g^\prime(1/u)}\\ & =\lim_{x\to+\infty}\frac{f^\prime(x)}{g^\prime(x)}.\quad (\text{replace} \frac{1}{u}=x)\end{aligned}\]
Notice that \(f^\prime(x)/g^\prime(x)\) is the derivative of 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.
To apply l’Hôpital’s rule:
Make sure the limit of \(f(x)/g(x)\) assumes the form 0/0 as \(x\to s\).
Differentiate \(f(x)\) and \(g(x)\) separately.
Find the limit of \(f^\prime(x)/g^\prime(x)\) 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.
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 type of error to apply 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
Find \[\lim_{t\to2}\frac{\sqrt{t+7}-3}{t-2}.\]
Solution 4
Because \[\left.\frac{\sqrt{t+7}-3}{t-2}\right|_{t=2}=\frac{\sqrt{2+7}-3}{t-2}=\frac{0}{0},\] we apply l’Hôpital’s rule: \[\begin{aligned} \lim_{t\to2}\frac{\sqrt{t+7}-3}{t-2} & \stackrel{H}{=}\lim_{t\to2}\frac{\dfrac{1}{2\sqrt{t+7}}}{1}\\ \\ & =\frac{\dfrac{1}{2\sqrt{2+7}}}{1}\\ & =\frac{1}{6}.\end{aligned}\]
Example 5
Find \[\lim_{x\to0}\frac{1-\cos x}{x^2}.\]
Solution 5
Note that here we have an indeterminate limit of type 0/0. So \[\lim_{x\to0}\frac{1-\cos x}{x^{2}}\stackrel{H}{=}\lim_{x\to0}\frac{-(-\sin x)}{2x}=\frac{1}{2}.\] For the last step, we could use l’Hôpital’s rule again.
Example 6
Find \[\lim_{x\to0}\frac{x-\sin x}{x-\tan x}.\]
Solution 6
Here we have an indeterminate limit of type \(0/0\). So
\[\lim_{x\to0}\frac{x-\sin x}{x-\tan x}\stackrel{H}{=}\lim_{x\to0}\frac{1-\cos x}{1-\sec^{2}x}\] Although the second limit is also indeterminate and we can apply l’Hôpital’s rule again, it is easier to directly evaluate the second limit: \[\begin{aligned} \frac{1-\cos x}{1-\sec^{2}x} & =\frac{1-\cos x}{1-\frac{1}{\cos^{2}x}}\\ \\ & =\frac{1-\cos x}{\frac{\cos^{2}x-1}{\cos^{2}x}}\\ \\ &=\frac{\cos^{2}x\ (1-\cos x)}{(\cos x-1)(\cos x+1)}\quad{(A^{2}-B^{2}=(A-B)(A+B))}\\ \\ & =\frac{-\cos^{2}x}{\cos x+1}.\end{aligned}\] Therefore: \[\lim_{x\to0}\frac{1-\cos x}{1-\sec^{2}x}=\lim_{x\to0}\frac{-\cos^{2}x}{\cos x+1}=-\frac{1}{1+1}=-\frac{1}{2}.\]
Example 7
Find \[\lim_{x\to1}\frac{\ln x}{x-1}.\]
Solution 7
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 8
Find \[\lim_{x\to0}\frac{x-\sin x}{\sinh x}.\]
Solution 8
Because \[\frac{x-\sin x}{\sinh x}=\frac{0-0}{0},\] we have
Because \[\lim_{x\to0}(\arcsin4x-2\arcsin2x)=0-2(0)=0\] and \[\lim_{x\to0}x^{3}=0\] we have an indeterminate limit of type \(0/0\) and we can apply l’Hôpital’s rule . Recall that \[(\arcsin x)’=\frac{1}{\sqrt{1-x^{2}}}.\] Thus, \[\lim_{x\to0}\frac{\arcsin4x-2\arcsin2x}{x^{3}}\stackrel{H}{=}\lim_{x\to0}\frac{\dfrac{4}{\sqrt{1-16x^{2}}}-2\dfrac{2}{\sqrt{1-4x^{2}}}}{3x^{2}}\] The second limit is still 0/0. Thus we can apply l’Hôpital’s rule again. To differentiate the numerator, it might be easier to write it as \[4(1-16x^{2})^{-\frac{1}{2}}-4(1-4x^{2})^{-\frac{1}{2}}.\] Therefore, \[\begin{aligned} &\lim_{x\to0}\frac{4(1-16x^{2})^{-\frac{1}{2}}-4(1-4x^{2})^{-\frac{1}{2}}}{3x^{2}} \\ & \stackrel{H}{=}\lim_{x\to0}\frac{\dfrac{4(-32\bcancel{x})}{-2}(1-16x^{2})^{-\frac{3}{2}}-\dfrac{4(-8\bcancel{x})}{-2}(1-4x^{2})^{-\frac{3}{2}}}{6\bcancel{x}}\\ & =\lim_{x\to0}\frac{64(1-16x^{2})^{-\frac{3}{2}}-16(1-4x^{2})^{-\frac{3}{2}}}{6}\\ & =\frac{64-16}{6}\\ & =8.\end{aligned}\]
Because \[\lim_{x\to\infty}e^{1/x^{2}}=e^{\lim_{x\to\infty}\frac{1}{x^{2}}}=e^{0}=1\] and \[\lim_{x\to\infty}\arctan x^{2}\stackrel{u=x^{2}}{=}\lim_{u\to\infty}\arctan u=\frac{\pi}{2},\] we have an indeterminate limit of type \(0/0\). Thus \[\lim_{x\to\infty}\frac{e^{1/x^{2}}-1}{2\arctan x^{2}-\pi}\stackrel{H}{=}\lim_{x\to\infty}\frac{\dfrac{d}{dx}\left(e^{1/x^{2}}-1\right)}{\dfrac{d}{dx}\left(2\arctan x^{2}-\pi\right)}.\] Because \[\frac{d}{dx}e^{u}=e^{u}\frac{du}{dx},\quad\frac{d}{dx}\arctan u=\frac{1}{1+u^{2}}\dfrac{du}{dx},\] we obtain \[\begin{aligned} \lim_{x\to\infty}\frac{\dfrac{d}{dx}\left(e^{1/x^{2}}-1\right)}{\dfrac{d}{dx}\left(2\arctan x^{2}-\pi\right)} & =\lim_{x\to\infty}\frac{\dfrac{d}{dx}(x^{-2})e^{1/x^{2}}}{2\dfrac{dx^{2}}{dx}\dfrac{1}{1+x^{4}}}\\ \\ & =\lim_{x\to\infty}\frac{-\cancel{2}x^{-3}e^{1/x^{2}}}{2(\cancel{2}x)\dfrac{1}{1+x^{4}}}\\ \\ & =\lim_{x\to\infty}\frac{1+x^{4}}{-2x^{4}}e^{1/x^{2}}\\ \\ & =\left(\lim_{x\to\infty}\frac{1+x^{4}}{-2x^{4}}\right)\left(\lim_{x\to\infty}e^{1/x^{2}}\right)\\ \\ & =\left(\lim_{x\to\infty}\frac{x^{4}}{-2x^{4}}\right)\left(e^{\lim_{x\to\infty}1/x^{2}}\right)\\ \\ & =\left(\frac{1}{-2}\right)(\underbrace{e^{0}}_{=1})\\ & =-\frac{1}{2}.\end{aligned}\]
Because \[\lim_{x\to+\infty}x^{-5/3}=\lim_{x\to+\infty}\frac{1}{x^{5/3}}=0\] and \[\begin{aligned} \lim_{x\to+\infty}\sin\left(\frac{1}{x}\right) & =\sin\left(\lim_{x\to+\infty}\frac{1}{x}\right)\\ & =\sin0=0,\end{aligned}\] we have an indeterminate limit of type \(0/0\). So l’Hôpital’s rule applies: \[\begin{aligned} \lim_{x\to+\infty}\frac{x^{-5/3}}{\sin(1/x)} & \stackrel{H}{=}\lim_{x\to+\infty}\frac{-\frac{5}{3}x^{-8/3}}{-\frac{1}{x^{2}}\cos\left(1/x\right)}\\ & =\lim_{x\to+\infty}\frac{-\frac{5}{3}x^{-2/3}}{\cos(1/x)}\\ & =\lim_{x\to+\infty}\frac{1}{\cos(1/x)}\left(\lim_{x\to+\infty}-\frac{5}{3x^{2/3}}\right)\\ & =\frac{1}{\cos0}(0)\\ & =0.\end{aligned}\]
Because \(\sin x\) and \(\ln x\) are continuous functions, we have \[\begin{aligned} \lim_{x\to-\infty}\sin\frac{1}{x} & =\sin\left(\lim_{x\to-\infty}\frac{1}{x}\right)\\ & =\sin0=0\end{aligned}\] and \[\begin{aligned} \lim_{x\to-\infty}\ln\frac{x+1}{x+2} & =\ln\left(\lim_{x\to-\infty}\frac{x+1}{x+2}\right)\\ & =\ln1=0.\end{aligned}\] So we have an indeterminate limit of type 0/0, and l’Hôpital’s rule applies: \[\begin{aligned} \lim_{x\to-\infty}\frac{\sin(1/x)}{\ln\frac{x+1}{x+2}} & \overset{H}{=}\lim_{x\to-\infty}\frac{-\frac{1}{x^{2}}\cos\frac{1}{x}}{\frac{1(x+2)-1(x+1)}{(x+2)^{2}}\frac{x+2}{x+1}}\\ \\ & =\lim_{x\to-\infty}\frac{-\frac{1}{x^{2}}\cos\frac{1}{x}}{\frac{1}{(x+2)}\frac{1}{(x+1)}}\\ \\ & =\lim_{x\to-\infty}\left(-\frac{(x+1)(x+2)}{x^{2}}\cos\frac{1}{x}\right)\\ & =-\left(\lim_{x\to-\infty}\frac{(x+1)(x+2)}{x^{2}}\right)\left(\lim_{x\to-\infty}\cos\frac{1}{x}\right)\\ & =-\left(\lim_{x\to-\infty}\frac{x^{2}}{x^{2}}\right)\cos\left(\lim_{x\to-\infty}\frac{1}{x}\right)\\ & =-(1)\cos(0)\\ & =-1.\end{aligned}\]
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.
Example 13
Can we apply l’Hôpital’s rule to find \[\lim_{x\to0}\frac{f(x)}{g(x)}\] if \(f(x)=x^{2}\sin\dfrac{1}{x}\) and \(g(x)=x\)?
Solution 13
In this example, \[f(x),g(x)\to0\text{ as }x\to0\] [Because \(-1\leq\sin\dfrac{1}{x}\leq1\), hence \(-x^{2}\leq x^{2}\sin\dfrac{1}{x}\leq x^{2}\), and by the Sandwich Theorem we have \(\lim_{x\to0}x^{2}\sin\frac{1}{x}=0\).]
So we deal with an indeterminate limit of type \(0/0\). If we attempt to apply l’Hôpital’s rule, we get \(g^\prime(x)=1\), \[\begin{aligned} f^\prime(x) & =2x\sin\frac{1}{x}+x^{2}\left(-\frac{1}{x^{2}}\right)\cos\frac{1}{x}\\ & =2x\sin\frac{1}{x}-\cos\frac{1}{x},\end{aligned}\] and \[\lim_{x\to0}\frac{f^\prime(x)}{g^\prime(x)}=\lim_{x\to0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right).\] However, this limit does not exist because \(2x\sin\frac{1}{x}\to0\) (by The Sandwich Theorem), but the limit of \(\cos\frac{1}{x}\) does not exist as it widly oscillates between \(-1\) and \(1\).
Remark that we CANNOT conclude that \(\lim_{x\to0}f(x)/g(x)\) does not exist. In fact, in this example, \(\lim_{x\to0}f(x)/g(x)\) DOES exist: \[\begin{aligned} \lim_{x\to0}\frac{f(x)}{g(x)} & =\lim_{x\to0}\frac{x^{2}\sin\frac{1}{x}}{x}\\ & =\lim_{x\to0}x\sin\frac{1}{x}=0.\end{aligned}\] In general, if \(\lim_{x\to s}f^\prime(x)/g^\prime(x)\) does not exist (and it is not equal to \(\pm\infty\)), l’Hôpital’s rule does not apply.
L’Hópital’s rule for the indeterminate form \(\pm\infty/\pm\infty\)
Theorem 2.(L’Hôpital’s Rule for \(\infty\)/\(\infty\)): Assume \[\lim_{x\to s}f(x)=+\infty,\text{ or }-\infty\] and \[{\displaystyle \lim_{x\to s}g(x)=+\infty}\text{ or }-\infty.\] If \[\lim_{x\to s}\frac{f^\prime(x)}{g^\prime(x)}=L,\] then \[\lim_{x\to s}\frac{f(x)}{g(x)}=L.\] Here \(s\) signifies \(a,a^{+},a^{-},-\infty\), or \(+\infty\), where \(a\in\mathbb{R}\). The Theorem remains valid if \(L\) is replaced by \(+\infty\) or \(-\infty\).
The proof of this theorem is discussed in more advanced books.
Because \[\lim_{x\to\frac{\pi}{2}^{+}}\tan5x=\lim_{x\to\frac{\pi}{2}^{+}}\tan x=-\infty,\] we have \[\begin{aligned} \lim_{x\to\frac{\pi}{2}^{+}}\frac{\tan5x}{\tan x} & \stackrel{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{5\sec^{2}5x}{\sec^{2}x}\\ &=\lim_{x\to\frac{\pi}{2}^{+}}\frac{5\cos^{2}x}{\cos^{2}5x}\qquad(0/0)\\ & \overset{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{\cancel{5(2)}(\bcancel{-}\sin x)\cos x}{\cancel{2(5)}(\bcancel{-}\sin5x)\cos5x}\\ & \begin{equation*}=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin2x}{\sin10x}\tag{using double angle formula}\end{equation*}\\ & \stackrel{H}{=}\lim_{x\to\frac{\pi}{2}^{+}}\frac{2\cos2x}{10\cos10x}\\ & =\frac{2\cos0}{10\cos0}\\ & =\frac{1}{5}.\end{aligned}\]
Example 15
Evaluate \[\lim_{x\to+\infty}\frac{\ln x}{x^{\alpha}},\] where \(\alpha>0\).
Solution 15
This is an indeterminate limit of type \(\infty/\infty\), and l’Hôpital’s rule applies: \[\begin{aligned} \lim_{x\to+\infty}\frac{\ln x}{x^{\alpha}} & \stackrel{H}{=}\lim_{x\to+\infty}\frac{\dfrac{1}{x}}{\alpha x^{\alpha-1}}\\ & =\lim_{x\to+\infty}\frac{1}{\alpha x^{\alpha}}\\ & =0,\end{aligned}\] because \(\alpha>0\).
The above example shows us that \(x^{\alpha}\) (\(\alpha>0\)) grows much faster than \(\ln x\).
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 \(0\cdot\pm\infty\)
We can apply l’Hôpital’s rule to evaluate the limits of the indeterminate form \(0\cdot(\pm\infty)\). Namely,
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 (Theorem 1 or 2).
Example 17
Evaluate \[\lim_{x\to\pi/2}(\sec3x\cos5x).\]
Solution 17
Because \[\begin{aligned} \lim_{x\to\pi/2}\sec3x & =\lim_{x\to\pi/2}\frac{1}{\cos3x}\\ & =\begin{cases} \lim_{x\to\frac{\pi}{2}^{-}}\frac{1}{\cos3x}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}+\infty & \text{(see the following fig)}\\ \lim_{x\to\frac{\pi}{2}^{+}}\frac{1}{\cos3x}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}-\infty & \text{(see the following fig)} \end{cases}\end{aligned}\] and \[\lim_{x\to\pi/2}\cos5x=\cos(5\pi/2)=0.\]Figure 3. (a) \(\cos x\to0^{+}\) as \(x\to\pi/2^{-}\) (b) \(\cos x\to0^{-}\) as \(x\to\pi/2^{+}\)
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 \(0/0\): \[\lim_{x\to\pi/2}\frac{\cos5x}{\cos3x}=\left[\frac{\cos(5\pi/2)}{\cos(3\pi/2)}=\frac{0}{0}\right]\]
Figure 4
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.
Figure 5
Example 18
Evaluate \[\lim_{x\to\pi/2}(\pi-2x)\tan x.\]
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).
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.
Figure 7
Example 19
Evaluate \[\lim_{x\to0^{+}}x\ln\sin x.\]
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).
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.
Figure 9
Example 20
Evaluate \[\lim_{x\to-\infty}xe^{x}.\]
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 \(y=xe^{x}\).
Figure 10
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:
We notice \[\lim_{x\to\pi/2^{-}}\sec x=\lim_{x\to\pi/2^{-}}\frac{1}{\cos x}\overset{\left[\frac{1}{0^{+}}\right]}{=}+\infty\]
\[\lim_{x\to\pi/2^{-}}\tan x=\lim_{x\to\pi/2^{-}}\frac{\sin x}{\cos x}\overset{\left[\frac{1}{0^{+}}\right]}{=}+\infty,\] and \[\lim_{x\to\pi/2^{+}}\sec x=\lim_{x\to\pi/2^{+}}\frac{1}{\cos x}\overset{\left[\frac{1}{0^{-}}\right]}{=}-\infty\]\[\lim_{x\to\pi/2^{+}}\tan x=\lim_{x\to\pi/2^{+}}\frac{\sin x}{\cos x}\overset{\left[\frac{1}{0^{-}}\right]}{=}-\infty,\] so in both cases, we have an indeterminate limit of type \(\infty-\infty\). Using a common denominator, we get \[\begin{aligned} \lim_{x\to\frac{\pi}{2}}\left(\sec x-\tan x\right) & =\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right)\\ & =\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{\cos x}\left[=\frac{0}{0}\right]\\ & \overset{H}{=}\lim_{x\to\frac{\pi}{2}}\frac{-\cos x}{-\sin x}\\ & =\frac{\cos(\pi/2)}{\sin(\pi/2)}\\ & =-\frac{0}{1}\\ & =0.\end{aligned}\]
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: \[\begin{aligned} \lim_{x\to0}\frac{x-\sin x}{x\sin x} & \overset{H}{=}\lim_{x\to0}\frac{1-\cos x}{\sin x+x\cos x}\left[=\frac{0}{0}\right]\\ & \overset{H}{=}\lim_{x\to0}\frac{+\sin x}{\cos x+\cos x-x\sin x}\\ & =\frac{\sin0}{2\cos0-0(0)}\\ & =0.\end{aligned}\]
Here we have a limit of the form \(\infty-\infty\) (if you consider \(x\to1^{+}\) and \(x\to1^{-}\) separately, you will get \(\infty-\infty\)). Using a common denominator, we get \[\begin{aligned} \lim_{x\to1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right) & =\lim_{x\to1}\frac{x-1-\ln x}{(x-1)\ln x}\\ & \begin{equation*}=\lim_{x\to1}\frac{x-1-\ln x}{x\ln x-\ln x}\tag{0/0}\end{equation*}\end{aligned}\] Now we can apply l’Hôpital’s rule: \[\begin{aligned} \lim_{x\to1}\frac{x-1-\ln x}{x\ln x-\ln x} & \begin{equation*}\overset{H}{=}\lim_{x\to1}\frac{1-\frac{1}{x}}{\ln x-\frac{x}{x}-\frac{1}{x}}\tag{still 0/0}\end{equation*}\\ & \overset{H}{=}\lim_{x\to1}\frac{\frac{1}{x^{2}}}{\frac{1}{x}+\frac{1}{x^{2}}}\\ & =\frac{1}{1+1}=\frac{1}{2}.\end{aligned}\]
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
If \(f\to0\) and \(g\to0\)
If \(f\to1\) and \(g\to+\infty\) or \(-\infty\)
If \(f\to+\infty\) or \(-\infty\) and \(g\to0\).
To evaluate such limits, let
\[y=f(x)^{g(x)}.\] Taking the natural logarithm of each side, \[\begin{aligned} \ln y & =\ln f(x)^{g(x)}\\ & =g(x)\ln f(x),\end{aligned}\] and then exponentiating the results, we obtain \[\underbrace{e^{\ln y}}_{=y}=e^{g(x)\ln f(x)}.\] Because \(e^{x}\) is a continuous function \[\lim_{x\to s}y=e^{\lim_{x\to s}\left(g(x)\ln f(x)\right)}.\] 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 \[y=e^{g(x)\ln f(x)}\stackrel{e^{+\infty}}{\to}+\infty.\]
Example 24
Evaluate \[\lim_{x\to0^{+}}x^{x}.\]
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 \[y\to1\text{ as }x\to0^{+}.\]Figure 11
Example 25
Evaluate \[\lim_{x\to+\infty}x^{1/x}.\]
Solution 25
Because \(1/x\to0\) as \(x\to+\infty\), we have an indeterminate limit of type \((+\infty)^{0}\). Method a: To find this limit, we can take the natural logorithm of \(x^{1/x}\) and follow similiar steps of the previous example, \[y=x^{1/x}\]\[\Rightarrow\ln y=\ln\left(x^{1/x}\right)=\frac{1}{x}\ln x\] Because \(\ln x\to+\infty\) as \(x\to+\infty\), \(\ln x/x\) assumes an indeterminate form \(\infty/\infty\), and l’Hôpital’s rule applies: \[\lim_{x\to+\infty}\ln y=\lim_{x\to+\infty}\frac{\ln x}{x}=\lim_{x\to+\infty}\frac{\frac{1}{x}}{1}=0.\] Therefore, \[\begin{aligned} \lim_{x\to+\infty}y & =\lim_{x\to+\infty}e^{\ln y}\\ & =e^{\lim_{x\to+\infty}\ln y}\\ & =e^{0}=1.\end{aligned}\]Method b: Put \(u=1/x\). Then \(u\to0^{+}\) as \(x\to+\infty\) and \[\begin{aligned} \lim_{x\to+\infty}x^{1/x} & =\lim_{u\to0^{+}}\left(\frac{1}{u}\right)^{u}\\ & =\frac{1}{\lim_{u\to0^{+}}u^{u}}\end{aligned}\] From the previous example, we know \(\lim_{u\to0^{+}}u^{u}=1\). So \[\lim_{x\to+\infty}x^{1/x}=\frac{1}{1}=1.\]
Example 26
Evaluate \[\lim_{x\to0}(1+\sin x)^{1/x}.\]
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 \[y\to e^{1}=e\text{ as }x\to0.\]Figure 12
Example 27
Evaluate \[\lim_{x\to0^{+}}(\cot x)^{\sin x}.\]
Solution 27
Because \[\sin x\to0^{+}\] and \[\cot x=\frac{\cos x}{\sin x}\stackrel{\left[\frac{1}{0^{+}}\right]}{\rightarrow}+\infty\] as \(x\to0^{+}\), we have an indeterminate limit of type \(\infty^{0}\). Let \[y=(\cot x)^{\sin x}\] thus \[\ln y=\sin x\ln\cot x.\quad\left[=0\cdot\infty\text{ as }x\to0^{+}\right]\]\[\begin{aligned} \lim_{x\to0^{+}}\sin x\ln\cot x & =\lim_{x\to0^{+}}\frac{\ln\cot x}{\frac{1}{\sin x}}\left[=\frac{+\infty}{+\infty}\right]\\ & \stackrel{H}{=}\lim_{x\to0^{+}}\frac{-(1+\cot^{2}x)\ \frac{1}{\cot x}}{-\frac{\cos x}{\sin^{2}x}}\\ & =\lim_{x\to0^{+}}\frac{-(1+\cot^{2}x)\ \frac{\sin x}{\cos x}}{-\frac{\cos x}{\sin^{2}x}}\quad (\cot x=\frac{\cos x}{\sin x})\\ & =\lim_{x\to0^{+}}\left((1+\cot^{2}x)\frac{\sin^{3}x}{\cos^{2}x}\right)\\ \\ & =\lim_{x\to0^{+}}\left(\frac{\overbrace{\cos^{2}x+\sin^{2}x}^{=1}}{\sin^{2}x}\frac{\sin^{3}x}{\cos^{2}x}\right)\quad (\text{expanding} 1+\cot^{2}x)\\ & =\lim_{x\to0^{+}}\frac{\sin x}{\cos x}=0.\end{aligned}\] Because \(\ln y\to0\) as \(x\to0^{+}\), \[y=e^{\ln y}\to e^{0}=1\text{ as }x\to0.\]
Use Equation (i) only when \(f(a)=g(a)=0\). If this condition is not satisfied, we cannot use this equation.
