Question

Let $f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{(\sin (x+h))^{\ln (x+h)}-(\sin x)^{\ln x}}{h}$, then $f\left(\frac{\pi}{2}\right)$ is

• A. equal to 0
• B. equal to 1
• C. $\ln \frac{\pi}{2}$
• D. non-existent

Answer 1

Answer: (A)

Let $g(x)=(\sin x)^{\ln x}=e^{\ln x \cdot \ln (\sin x)}$
$f(x)=g^{\prime}(x)=(\sin x)^{\ln x}\left[\cot x(\ln x)+\frac{\ln (\sin x)}{x}\right]$
Hence, $f \left(\frac{\pi}{2}\right)= g ^{\prime}\left(\frac{\pi}{2}\right)=1(0+0)=0$