Question

If the function
$f\left(x\right)=\frac{\tan \left(\tan x\right)-\sin \left(\sin x\right)}{\tan x-\sin x}(x\ne 0)$,
is continous at $x=0$, then find the value of $f\left(0\right)$

• A. $4$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

Answer: (B)

$1$

We have,

$f\left(x\right)=\frac{\tan \left(\tan x\right)-\sin \left(\sin x\right)}{\tan x-\sin x}$

Since, $x=0$

Therefore,

${lim}_{x\to 0}\frac{\tan \left(\tan x\right)-\sin \left(\sin x\right)}{\tan x-\sin x}$

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

Hence, this is the answer.