Question

Let $f ( x )=\frac{ e ^{\tan x }- e ^{ x }+\ln (\sec x +\tan x )- x }{\tan x - x }$ be a continuous function at $x =0$. The value of $f (0)$ equals

• A. $\frac{1}{0}$
• B. $\frac{2}{2}$
• C. $\frac{3}{2}$
• D. 2

Answer 1

Answer: (C)

For continuity of f at $x = 0,$ we have
$k=f(0)=\underset{x \rightarrow 0}{\text{Lim}} f(x)=
\underset{x \rightarrow 0}{\text{Lim}} \frac{e^{\tan x}-e^x}{\tan x-x}+\underset{x \rightarrow 0}{\text{Lim}} \frac{\ln (\sec x+\tan x)-x}{\left(\frac{\tan x-x}{x^3}\right) x^3}$
$=\underset{x \rightarrow 0}{\text{Lim}} \frac{e^x\left(e^{\tan x-x}-1\right)}{\tan x-x}+3 \underset{x \rightarrow 0}{\operatorname{Lim}} \frac{\ln (\sec x-\tan x)-x}{x^3}=1+3\underset{x \rightarrow 0}{\text{Lim}} \frac{\sec x-1}{3 x^2} \text { (Using L.H. Rule) } $
$=1+\frac{1}{2}=\frac{3}{2} $