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
$k=f(0)=\underset{x\to 0}{\mathrm{Lim}}f(x)=<br/>\underset{x\to 0}{\mathrm{Lim}}\frac{e^{\tan x}-e^x}{\tan x-x}+\underset{x\to 0}{\mathrm{Lim}}\frac{\ln (\sec x+\tan x)-x}{\left(\frac{\tan x-x}{x^3}\right)x^3}$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{e^x\left(e^{\tan x-x}-1\right)}{\tan x-x}+3\underset{x\to 0}{\mathrm{Lim}}\frac{\ln (\sec x-\tan x)-x}{x^3}=1+3\underset{x\to 0}{\mathrm{Lim}}\frac{\sec x-1}{3x^2}\text{ (Using L.H. Rule) }$
$=1+\frac{1}{2}=\frac{3}{2}$