Question

If $\underset{x\to 0}{lim}\frac{a\tan 3x+(1-\cos 2x)}{x+\sin x+\tan x}=1$, then the value of $a$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

$\underset{x\to 0}{lim}\frac{a\tan 3x+(1-\cos 2x)}{x+\sin x+\tan x}=1\Rightarrow \underset{x\to 0}{lim}\frac{a\tan 3x+2{\sin }^{2}x}{x+\sin x+\tan x}=1$
Dividing the numerator and denominator by $x$,
$\Rightarrow \underset{x\to 0}{lim}\frac{3a\frac{\tan 3x}{3x}+2\frac{{\sin }^{2}x}{x}}{1+\frac{\sin x}{x}+\frac{\tan x}{x}}=1\Rightarrow \frac{3a+0}{1+1+1}=1$
$\Rightarrow \frac{3a}{3}=1\Rightarrow a=1$