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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0^+} (\cot x)^{\sin x}. \]

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*}

(Also, note that in the fourth line where we write \lim (A - B) = \lim A - \lim B, we are justified in doing this since both limits exist.)

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