OBJECTIVES

When you finish this section, you should be able to:

1. Identify indeterminate forms of the type $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }$ (p. 298)
2. Use L’Hôpital’s Rule to find a limit (p. 299)
3. Find the limit of an indeterminate form of the type $0\cdot \infty$, $\infty -\infty$, $0^{0}$, $1^{\infty }$, or $\infty ^{0}$ (p. 303)

NEED TO REVIEW?

Properties of limits are discussed in Section 1.2, pp. 87-88.

NEED TO REVIEW?

Limits at infinity and infinite limits are discussed in Section 1.5, pp. 117-125.

NOTE

The word “indeterminate” conveys the idea that the limit cannot be found without additional work.

DEFINITION Indeterminate Form at $c$ of the Type $\dfrac{0}{0}$ or the Type $\dfrac{\infty }{\infty }$

If the functions $f$ and $g$ are each defined in an open interval containing the number $c,$ except possibly at $c$, then the quotient $\dfrac{f(x) }{g(x) }$ is called an indeterminate form at $c$ of the type $\dfrac{0}{ 0}$ if $$\lim\limits_{x\rightarrow c}f(x) =0\qquad \hbox{and} \qquad \lim\limits_{x\rightarrow c}g(x) =0$$

and an indeterminate form at $c$ of the type $\dfrac{\infty }{\infty }$ if $$\lim\limits_{x\rightarrow c}f(x) =\pm \infty\qquad \hbox{and}\qquad \lim\limits_{x\rightarrow c}g(x) =\pm \infty$$

NOW WORK

Problem 7.

THEOREM L’Hôpital’s Rule

Suppose the functions $f$ and $g$ are differentiable on an open interval $I$ containing the number $c$, except possibly at $c,$ and $g^\prime (x) \neq 0$ for all $x\neq c$ in $I$. Let $L$ denote either a real number or $\pm \infty$, and suppose $\dfrac{f(x) }{g(x) }$ is an indeterminate form at $c$ of the type $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$. If $\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) }=L$, then $\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }=L.$

Proof

Suppose $\lim\limits_{x\rightarrow c}f(x) =0$ and $\lim\limits_{x\rightarrow c}g(x) =0.$ Then $f( c) =0$ and $g( c) =0.$ Since both $f^\prime$ and $g^\prime$ are continuous at $c$ and $g^\prime ( c) \neq 0,$ we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) } = \dfrac{\lim\limits_{x\rightarrow c}f^\prime (x) }{ \lim\limits_{x\rightarrow c}g^\prime (x) }= \dfrac{ f^\prime ( c) }{g^\prime ( c) }=\dfrac{ \lim\limits_{x\rightarrow c}\dfrac{f(x) -f( c) }{x-c}}{\lim\limits_{x\rightarrow c}\dfrac{g(x) -g( c) }{x-c}}\\[-21.8pt] &&\hspace{5.6pc}\underset{\color{#0066A7}{{\hbox{Quotient}}\atop{\hbox{Property}}}} {\color{#0066A7}{{\uparrow}}} \quad\hspace{28pt} \underset{ \color{#0066A7}{{\hbox{$f^\prime ,\,\,g^\prime$} }\atop{\hbox{continuous}}}} {\color{#0066A7}{\uparrow}} \quad\kern-6pt \underset{\color{#0066A7}{\hbox{Definition of}}\atop\color{#0066A7}{\hbox {a derivative}}}{\hspace{-1pc}{\hspace{-12pt}{\color{#0066A7}{\uparrow}}\hspace{12pt}}}\\[5pt] &&\hspace{7pc}=\lim\limits_{x\rightarrow c}\dfrac{\dfrac{f(x) -f( c) }{x-c}}{\dfrac{g(x) -g( c) }{x-c}}=\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) } \\[-21.7pt] &&\hspace{19pc}\underset{\color{#0066A7}{\hbox{$f( c) =0$}}\atop\color{#0066A7}{\hbox{$g( c) =0$}}}{\color{#0066A7}{\uparrow}} \end{eqnarray*}$$

ORIGINS

Guillaume François de L’Hôpital (1661–1704) was a French nobleman. When he was 30, he hired Johann Bernoulli to tutor him in calculus. Several years later, he entered into a deal with Bernoulli. L’Hôpital paid Bernoulli an annual sum for Bernoulli to share his mathematical discoveries with him but no one else. In 1696 L’Hôpital published the first textbook on differential calculus. It was immensely popular, the last edition being published in 1781 (seventy-seven years after L’Hôpital’s death). The book included the rule we study here. After L’Hôpital’s death, Johann Bernoulli made his deal with L’Hôpital public and claimed that L’Hôpital’s textbook was his own material. His position was dismissed because he often made such claims. However, in 1921, the manuscripts of Bernoulli’s lectures to L’Hôpital were found, showing L’Hôpital’s calculus book was indeed largely Johann Bernoulli’s work.

Steps for Finding a Limit Using L’Hôpital’s Rule

Step 1 Check that $\dfrac{f}{g}$ is an indeterminate form at $c$ of the type $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$. If it is not, do not use L’Hôpital’s Rule.

Step 2 Differentiate $f$ and $g$ separately.

Step 3 Find $\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) }$. This limit is equal to $\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }$, provided the limit is a number or $\infty$ or $-\infty$.

Step 4 If $\dfrac{f^\prime }{g^\prime }$ is an indeterminate form at $c$, repeat the process.

Find $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}$.

Solution We follow the steps for finding a limit using L’Hôpital’s Rule.

Step 1 Since $\lim\limits_{x\rightarrow 0}\tan x=0$ and $\lim\limits_{x\rightarrow 0}( 6x) =0$, the quotient $\dfrac{\tan x }{6x}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0}$.

Step 2 $\dfrac{d}{\textit{dx}}\tan x=\sec ^{2}x$ and $\dfrac{d}{\textit{dx}}\left( 6x\right) =6$.

Step 3 $\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}\tan x}{\dfrac{d}{dx}\left( 6x\right) }=$ $\lim\limits_{x\rightarrow 0}\dfrac{\sec ^{2}x}{6}=\dfrac{1}{6}.$

It follows from L’Hôpital’s Rule that $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}=\dfrac{1}{6}$.

CAUTION

When using L’Hôpital’s Rule, we find the derivative of the numerator and the derivative of the denominator separately. Be sure not to find the derivative of the quotient. Also, using L’Hôpital’s Rule to find a limit of an expression that is not an indeterminate form may result in an incorrect answer.

Find $\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}$.

Solution We use the steps for finding a limit using L’Hôpital’s Rule.

Step 1 Since $\lim\limits_{x\rightarrow 0}\left( \sin x-x\right) =0$ and $\lim\limits_{x\rightarrow 0}x^{2}=0$, the expression $\dfrac{\sin x-x}{x^{2}}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0}.$

Steps 2 and 3 We use L’Hôpital’s Rule. $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( \sin x-x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\cos x-1}{2x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0} \dfrac{\cos x-1}{x}\\ &&\hspace{8pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*}$$

Since $\lim\limits_{x\rightarrow 0}( \cos x-1) =0$ and $\lim\limits_{x\rightarrow 0}x=0$, the expression $\dfrac{\cos x-1}{x}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0}$. So, we use L’Hôpital’s Rule again. $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{\cos\;x-1}{x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}( \cos\;x-1) }{\dfrac{d}{\textit{dx}}x}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{-\!\sin\;x}{1}=0\\[-1.8pc] &&\hspace{17pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }} } \end{eqnarray*}$$

NOW WORK

Problem 29.

Find:

Solution (a) Since $\lim\limits_{x\rightarrow \infty }\ln x=\infty$ and $\lim\limits_{x\rightarrow \infty }x=\infty$, $\dfrac{\ln x}{x}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hôpital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x} = \lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{1}{x}}{1}=\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}=0\\[-1.8pc] &&\hspace{3.35pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*}$$

(b) $\lim\limits_{x\rightarrow \infty }x=\infty$ and $\lim\limits_{x\rightarrow \infty }e^{x}=\infty$, so $\dfrac{x}{e^{x}}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hôpital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{x}{e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{1}{e^{x}}=0\\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*}$$

(c) From (b), we know that $\dfrac{e^{x}}{x}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hô pital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}e^{x}}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{e^{x}}{1}=\infty \\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{ \begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array} }{{\uparrow }}} \end{eqnarray*}$$

NOW WORK

Problem 35.

Find $\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x}$.

Solution $\dfrac{\tan x-\!\sin x}{x^{2}\tan x}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0}$. We simplify the expression before using L’Hôpital’s Rule. Then it is easier to find the limit. $$\begin{eqnarray*} \dfrac{\tan x-\!\sin x}{x^{2}\tan x}&=&\dfrac{\dfrac{\sin x}{\cos x}-\!\sin x}{x^{2}\cdot \dfrac{\sin x}{\cos x}}=\dfrac{\dfrac{\sin x-\!\sin x\cos x}{\cos x}}{\dfrac{x^{2}\sin x}{\cos x}}=\dfrac{\sin x\left( 1-\cos x\right) }{x^{2}\sin x}=\dfrac{1-\cos x}{x^{2}} \end{eqnarray*}$$

Since $\dfrac{1-\cos x}{x^{2}}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0},$ we use L’Hôpital’s Rule.

$$\begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x} =\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( 1-\cos x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\sin x}{2x}\\[-20.7pt] &&\hspace{16.7pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\uparrow }}}\\[6pt] &&\hspace{9.7pc}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=\dfrac{1}{2}\qquad \color{#0066A7}{{\hbox{$\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x} =1$}} } \end{eqnarray*}$$

CAUTION

Be sure to verify that a simplified form is an indeterminate form at $c$ before using L’Hôpital’s Rule.

Find $\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}$.

Solution Since $\lim\limits_{x\rightarrow 0^{+}}\cot x=\infty$ and $\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty$, $\dfrac{\cot x}{\ln x}$ is an indeterminate form at $0^{+}$ of the type $\dfrac{\infty }{\infty }$. Using L’Hôpital’s Rule, we find $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\cot x}{\dfrac{d}{\textit{dx}}\ln x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\csc ^{2}x}{\dfrac{1}{x}}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{x}{ \sin ^{2}x}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}\sin ^{2}x}\\[-21pt] &&\hspace{4pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \ \hspace{10.5pc} \color{#0066A7}{\underset{\hbox{$\csc ^{2}x=\dfrac{1}{\sin^{2}x}$}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \, \hspace{4.5pt}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{7pc}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{2\sin x\cos x}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\cos x}\right)\\[7pt] &&\hspace{7pc}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) ( 1) =-\infty \end{eqnarray*}$$

NOW WORK

Problem 61.

NOTE

It might be tempting to argue that $0\cdot \infty$ is $0$, since “anything” times $0$ is $0$, but $0\cdot \infty$ is not a product of numbers. That is, $0\cdot \infty$ is not “zero times infinity” it symbolizes “a quantity tending to zero” times “a quantity tending to infinity.”

Find:

Solution (a) Since $\lim\limits_{x\rightarrow 0^{+}}x=0$ and $\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty$, then $x\ln x$ is an indeterminate form at $0^{+}$ of the type $0\cdot \infty$. We change $x\ln x$ to an indeterminate form of the type $\dfrac{\infty }{\infty }$ by writing $x\ln x= \dfrac{\ln x}{\dfrac{1}{x}}$ and using L’Hôpital’s Rule. $$\begin{eqnarray*} \lim\limits_{x\rightarrow 0^{+}}( x\ln x) &=&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\ln x}{\dfrac{1}{x}}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}\dfrac{1}{x}} =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}}= \lim\limits_{x\rightarrow 0^{+}}( -x) =0\\[-20.9pt] &&\hspace{16pt}\quad\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}}\qquad\hspace{4.6pc} \hspace{3.7pc}\color{#0066A7}{\underset{\hbox{Simplify}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}} \end{eqnarray*}$$

(b) Since $\lim\limits_{x\rightarrow \infty }\;x=\infty$ and $\lim\limits_{x\rightarrow \infty }\;\sin \dfrac{1}{x}=0$, then $x\sin \dfrac{1}{x}$ is an indeterminate form at $\infty$ of the type $0\cdot \infty$. We change $x\;\sin \dfrac{1}{x}$ to an indeterminate form of the type $\dfrac{0}{0}$ byeak writing $$\begin{eqnarray*} \lim\limits_{x\rightarrow \infty }x\;\sin \dfrac{1}{x}&=&\lim\limits_{x\rightarrow \infty } \dfrac{\sin \dfrac{1}{x}}{\dfrac{1}{x}}=\lim\limits_{t\rightarrow 0^{+}}\dfrac{\sin\;t}{t}=1 \\[-1.75pc] &&\qquad\hspace{26.1pt}\color{#0066A7}{\underset{{\hbox {Let} \;t=\dfrac{1}{x}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}} \end{eqnarray*}$$

NOTE

Notice in the solution to (b) that we did not need to use L’Hôpital’s Rule.

NOW WORK

Problem 45.

Find $\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin x}\right)$.

Solution Since $\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{x} =\infty$ and $\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}=\infty$, then $\dfrac{1}{x}-\dfrac{1}{\sin\;x}$ is an indeterminate form at $0^{+}$ of the type $\infty -\infty.$ We rewrite the difference as a single fraction. $$\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x}$$

Then, $\dfrac{\sin\;x-x}{x\sin\;x}$ is an indeterminate form at $0^{+}$ of the type $\dfrac{0}{0}$. Now we can use L’Hôpital’s Rule. $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\left(\sin\;x-x\right) }{\dfrac{d}{\textit{dx}}\left(x\;\sin x\right)}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cos\;x-1}{x\cos\;x+\sin\;x}\\[-20.7pt] &&\qquad\hspace{14.5pc}\color{#0066A7}{\underset{{\hbox{L'H$\hat{\rm o}$pital's Rule}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}} \left(\cos\;x-1\right) }{\dfrac{d}{\textit{dx}}\left(x\;\cos\;x+\;\sin\;x\right) } =\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\!\sin\;x}{\left(-x\;\sin\;x+\cos\; x\right) +\cos\;x}\\[-20.8pt] &&\hspace{3.5pc}\color{#0066A7}{\underset{\hbox{Type $\dfrac{0}{0}$; use L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\[-6pt] &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x}{x\sin\;x-2\cos\;x}=\dfrac{0}{-2}=0 \end{eqnarray*}$$

NOW WORK

Problem 47.

NEED TO REVIEW?

Properties of logarithms are discussed in Appendix A.1, pp. A-10 to A-11.

Steps for Finding ${\lim\limits_{x\rightarrow c} [\ f(x) ] ^{g(x)}}$ when ${[\ f(x)] ^{g(x)}}$ is an Indeterminate Form at $c$, of the Type ${1^{\infty}}$, ${0^{0}}$, or ${\infty ^{0}}$

Step 1 Let $y=[ f(x) ] ^{g(x)}$ and take the natural logarithm of each side, obtaining $\ln y=g(x)\;\ln f(x) .$

Step 2 Find $\lim\limits_{x\rightarrow c}\;\ln y$.

Step 3 If $\lim\limits_{x\rightarrow c}\;\ln y=L$, then $\lim\limits_{x\rightarrow c}y=e^{L}$.

Find $\lim\limits_{x\rightarrow 0^{+}}x^{x}$

Solution The expression $x^{x}$ is an indeterminate form at $0^{+}$ of the type $0^{0}.$ We follow the steps for finding $\lim\limits_{x\rightarrow c}[ f(x) ] ^{g(x) }$.

Step 1 Let $y=x^{x}.$ Then $\ \ln y=x\ln x.$

Step 2 $\lim\limits_{x\rightarrow 0^{+}}\;\ln y=\lim\limits_{x\rightarrow 0^{+}}( x\;\ln x) =0$ [from Example 7(a)].

Step 3 Since $\lim\limits_{x\rightarrow 0^{+}}\;\ln y=0$, $\lim\limits_{x\rightarrow 0^{+}}y=e^{0}=1$.

NOW WORK

Problem 51.

Find $\lim\limits_{x\rightarrow 0^{+}}( 1+x) ^{1/x}$.

Solution The expression $( 1+x) ^{1/x}$ is an indeterminate form at $0^{+}$ of the type $1^{\infty }$.

Step 1 Let $y=( 1+x) ^{1/x}.$ Then $\ln y=\dfrac{1}{x}\;\ln (1+x).$

Step 2 $$\begin{array}{l} &\lim\limits_{x\rightarrow 0^{+}}\;\ln y=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\ln ( 1+x) }{x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}\;\ln \left( 1+x\right) } {\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{1+x}}{1}=1\\ &\hspace{10pc}\hspace{-.7pt}\color{#0066A7}{\underset{\hbox{Type $\dfrac{0}{0}$; use L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ \end{array}$$

Step 3 Since $\lim\limits_{x\rightarrow 0^{+}}\ln y=1$, $\lim\limits_{x\rightarrow 0^{+}}y=e^{1}=e$.

NOW WORK

Problem 85.