Question

$f\left(x\right)=\frac{\tan x-e^x+\ln (\sec x+\tan x)}{\tan x}$ is continuose at $x=0$, then $f\left(0\right)$ equals

• A. $\frac{1}{2}$
• B. $\frac{19}{18}$
• C. $\frac{2}{3}$
• D. $-\infty$

Answer 1

The correct option is D $-\infty$

Consider the function,

$f\left(x\right)=\frac{\tan x-e^x+\ln (\sec x+\tan x)}{\tan x}$

The objective is to find the value of the function when $x=0$

$f\left(0\right)=\frac{\tan 0-e^0+\ln \left(\sec 0+\tan 0\right)}{\tan 0}$

$=\frac{0-1+0}{0}$

$=-\infty$

Hence, the value of the function is $f\left(0\right)=-\infty$