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