Question

$\underset{x\to 0}{lim}\left[\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}\right]$

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

Answer 1

The correct option is A ${\log }_{e}a$

$\underset{x\to 0}{lim}\left[\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}\right]$

$=\underset{x\to 0}{lim}\left[\frac{(a^{\tan x}-1)-(a^{\sin x}-1)}{\tan x-\sin x}\right]$

$=\underset{x\to 0}{lim}\left[\frac{(\frac{a^{\tan x}-1}{\tan x})\tan x-(\frac{a^{\sin x}-1}{\sin x})\sin x}{\tan x-\sin x}\right]$

we know that $\underset{x\to 0}{lim}\left[\frac{a^x-1}{x}\right]$=${\log }_{e}a$

$=\underset{x\to 0}{lim}\left[\frac{{\log }_{e}a\times \tan x-{\log }_{e}a\times \sin x)}{\tan x-\sin x}\right]$

$={\log }_{e}a$

Alternate:
$\underset{x\to 0}{lim}\left[\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}\right]=\underset{x\to 0}{lim}{a}^{\sin x}\left[\frac{a^{(\tan x-\sin x)}-1}{\tan x-\sin x}\right]=\ln a$