\[lim_{x \rightarrow \infty} (1+x)^{ \frac{1}{x}}\]
take logs.\[ln((1+x)^{ \frac{1}{x}})= \frac{1}{x} ln(1+x)= \frac{ln(1+x)}{x}\]
.
This is of Indeterminate Form as \[x \rightarrow \infty\]
(it equals \[\frac{ \infty}{ \infty }\]
)so use l'Hospital's Rule.\[\begin{equation} \begin{aligned} lim_{x \rightarrow \infty} \frac{ln(1+x)}{x} &= lim_{ x \rightarrow \infty} \frac{\frac{d}{dx} (ln(1+x))}{\frac{d}{dx}(x)} \\ &= lim_{ x \rightarrow \infty} \frac{1/(1+x)}{1} \\ &= 0 \end{aligned} \end{equation}\]
.Then
\[lim_{x \rightarrow \infty} (1+x)^{ \frac{1}{x}}=e^0=1\]
.