Evaluating Limits By Taking Logs

To evaluate  
\[lim_{x \rightarrow 0} (1+x)^{cotx}\]
  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  
\[lim_{x \rightarrow 0} (1+x)^{cotx}=e^1=e \]
.

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