Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{\theta\to0^+}(\sin\theta)^{\tan\theta}=1.$$
To solve this limit we will transform the expression under the limit. Since the natural exponential function and the natural logarithmic function are inverse to each other we have that $a=e^{\ln a}.$ This gives $(\sin\theta)^{\tan\theta}=e^{\ln(\sin\theta)^{\tan\theta}}=e^{\tan\theta\ln\sin\theta}$ where we used the logarithmic rule $\ln b^a=a\ln b$. Also because the exponential function is continuous it can exchange places with the limit so we get $$L=\lim_{\theta\to0^+}(\sin\theta)^{\tan\theta}=\lim_{\theta\to0^+}e^{\tan\theta\ln\sin\theta}=e^{\lim_{\theta\to0^+}\tan\theta\ln\sin\theta}=e^{\lim_{\theta\to0^+}\frac{\ln\sin\theta}{\frac{1}{\tan\theta}}}=e^l,$$ where we have denoted $l=\lim_{\theta\to0^+}\frac{\ln\sin\theta}{\frac{1}{\tan\theta}}.$ Let's solve this limit. "LR" will stand for "Apply L'Hopital's rule." $$l=\lim_{\theta\to0^+}\frac{\ln\sin\theta}{\frac{1}{\tan\theta}}=\left[\frac{\ln\sin 0^+}{\frac{1}{\tan 0^+}}\right]=\left[\frac{\ln0^+}{\frac{1}{0^+}}\right]=\left[\frac{-\infty}{\infty}\right][\text{LR}]=\lim_{\theta\to0^+}\frac{(\ln\sin\theta)'}{\left(\cot\theta\right)'}=\lim_{\theta\to0^+}\frac{\frac{1}{\sin\theta}(\sin\theta)'}{-\frac{1}{\sin^2\theta}}=\lim_{\theta\to0^+}\frac{\cos\theta}{-\frac{1}{\sin\theta}}=-\lim_{\theta\to0^+}\sin\theta\cos\theta=-\sin0\cos0=0.$$ Putting this into the initial limit we get $$L=e^l=e^0=1.$$