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