Question

Evaluate $\underset{x\to 0}{lim}\frac{a^{\sin x}-1}{\sin x}$.

Answer 1

$\underset{x\to 0}{lim}\frac{a^{\sin x}-1}{\sin x}$.

We have $\underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a$.

Using the above we have

$\underset{x\to 0}{lim}\frac{a^{\sin x}-1}{\sin x}={\log }_{e}a$.