Evaluate the limit.
![\[ \lim_{x \to 0^+} (\cot x)^{\sin x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-248e0351cd535da8831311b51899cc46_l3.png "Rendered by QuickLaTeX.com")
We use the definition and continuity of the exponential to rewrite things a bit, and then we apply L’Hopital’s rule.
![\begin{align*} \lim_{x \to 0^+} (\cot x)^{\sin x} &= \lim_{x \to 0^+} e^{ \sin x \log (\cot x) } \\[9pt] &= \exp \left( \lim_{x \to 0^+} \sin x \log \left(\frac{\cos x}{\sin x} \right) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \left( \sin x \log (\cos x) - \sin x \log (\sin x) \right) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} ( \sin x \log (\cos x)) - \lim_{x \to 0^+} ( \sin x \log (\sin x)) \right) \\[9pt] &= \exp \left( 0 - \lim_{x \to 0^+} \frac{\log (\sin x)}{\frac{1}{\sin x}} \right) \\[9pt] &= \exp \left(\lim_{x \to 0^+} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin^2 x}} \right) \\[9pt] &= \exp \left(\lim_{x \to 0^+} \sin x \right) \\[9pt] &= e^0 = 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-14348432d036d31af78bef8b117d435c_l3.png "Rendered by QuickLaTeX.com")
(Also, note that in the fourth line where we write 
, we are justified in doing this since both limits exist.)
It is not necessary to do L-hopital here. Just make a substitution t = sin(x) in the second limit, and apply Example 2.