Question

${lim}_{x\to 1}\frac{a^{x-1}-1}{sin\pi x}$

Answer 1

Consider the given function:

$f\left(x\right)={lim}_{x\to 1}\frac{a^{x-1}-1}{\sin \pi x}$

$weknowthat\left(\frac{0}{0}\right)form$

$={lim}_{x\to 1}\frac{a^{x-1}.l0g-0}{\left(\cos \pi x\right).\pi }$

$={lim}_{x\to 1}\frac{\log a.a^{x-1}}{\pi \cos \pi x}$

$=\frac{\log a.a^{1-1}}{\pi \cos \pi }$

$=\frac{\log a}{\pi }$

Hence this is the answer.