Đáp án:
\(\lim\limits_{x\to \tfrac{\pi}{2}}(\sin x)^{\tan x}= 1\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to \tfrac{\pi}{2}}(\sin x)^{\tan x}\\
= \lim\limits_{x\to \tfrac{\pi}{2}}e^{\displaystyle{\ln(\sin x)^{\tan x}}}\\
= e^{\displaystyle{\lim\limits_{x\to \tfrac{\pi}{2}}(\tan x\ln\sin x)}}\\
= e^{\displaystyle{\lim\limits_{x\to \tfrac{\pi}{2}}\left(\sin x\cdot \dfrac{\ln\sin x}{\cos x}\right)}}\\
= e^{\displaystyle{\lim\limits_{x\to \tfrac{\pi}{2}}\sin x\cdot \lim\limits_{x\to \tfrac{\pi}{2}}\dfrac{\ln\sin x}{\cos x}}}\\
= e^{\displaystyle{\sin\dfrac{\pi}{2}\cdot \lim\limits_{x\to \tfrac{\pi}{2}}\left(-\dfrac{\cos x}{\sin^2x}\right)}}\qquad \text{(l'Hôpital)}\\
= e^{\displaystyle{-\dfrac{\cos\dfrac{\pi}{2}}{\sin^2\dfrac{\pi}{2}}}}\\
= e^0\\
= 1
\end{array}\)
