\[lim_{x \rightarrow 0} (1+x)^{cotx}\]
take logs.\[ln((1+x)^{cotx})= cotx ln(1+x)= \frac{ln(1+x)}{tanx}\]
.
This is of Indeterminate Form as \[x \rightarrow 0\]
(it equals \[\frac{0}{0}\]
)so use l'Hospital's Rule.\[\begin{equation} \begin{aligned} lim_{x \rightarrow 0} \frac{ln(1+x)}{tanx} &= lim_{ x \rightarrow 0} \frac{\frac{d}{dx} (ln(1+x))}{\frac{d}{dx}(tanx)} \\ &= lim_{ x \rightarrow 0} \frac{1/(1+x)}{sec^2x} \\ &= 1 \end{aligned} \end{equation}\]
Then
\[lim_{x \rightarrow 0} (1+x)^{cotx}=e^1=e \]
.