Question

The value of $\underset{x\to 0}{lim}\frac{e^x+e^{-x}-2}{x^2}$ equals

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

Answer 1

Answer: (A)

$1$

${lim}_{x\to 0}\frac{e^x+e^{-x}-2}{x^2}$

on diffrentiating using L- Hospital rule

${lim}_{x\to 0}\frac{\frac{d}{dx}(e^x+e^{-x}-2)}{\frac{d}{dx}\left(x^2\right)}={lim}_{x\to 0}\frac{e^x-e^{-x}}{2x}$

again using L Hospital rule we have

$\Rightarrow {lim}_{x\to 0}\frac{e^x+e^{-x}}{2}$

On applying limit

$\Rightarrow \frac{1+1}{2}=1$