Question

$\underset{x\to 0}{\lim }\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}$ is equal to $(a>0)$

• A. ${\log }_{e}a$
• B. 1
• C. 0
• D. e

Answer 1

Answer: (A)

We have,
$\underset{x\to 0}{\lim }\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}=\underset{x\to 0}{\lim }{a}^{\sin x}\left(\frac{a^{\tan x-\sin x}-1}{\tan x-\sin x}\right)$
$={\log }_{e}a$
$\left[\because \underset{x\to 0}{\lim }\frac{a^x-1}{x}={\log }_{e}a\right]$