Calculus: Early Transcendentals (2nd Edition)

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