Question

The value of $\underset{x\to 0}{lim}\frac{e^x-(1+x)}{x^2}$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{2}$

Using exponential series, $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+.......$
$\underset{x\to 0}{lim}\frac{e^x-(1+x)}{x^2}$
$=\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+.......-(1+x)}{x^2}$
$={lim}_{x\to 0}(\frac{1}{2!}+\frac{x}{3!}+.......\text{higher order terms})=\frac{1}{2!}=\frac{1}{2}$