Question

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

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

Answer 1

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

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

on expanding $e^x$

$\underset{x\to 0}{lim}\frac{(1+x+\frac{x^2}{2!}.....)-(1+x)}{x^2}$

$\underset{x\to 0}{lim}(1+x)+$ $\frac{(\frac{x^2}{2!}+\frac{x^3}{3!}+....)-(1+x)}{x^2}$

$=\frac{1}{2!}+0+0=\frac{1}{2}$.