Question

$\displaystyle \lim_{x \to 0} \frac{a^{\tan \, x} - a^{\sin \, x}}{\tan \, x - \sin \, x}$ is equal to $(a > 0)$

• A. $\log_e a$
• B. 1
• C. 0
• D. e

Answer 1

Answer: (A)

$\log_e a$

Answer 2

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