Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{e^x-x-1}{2}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{e^x-x-1}{2}$

Exponential expansion is given by
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$
$\Rightarrow e^x-x-1=\frac{x^2}{2!}+\frac{x^3}{3!}+.....$

$\underset{x\to 0}{lim}\frac{e^x-x-1}{2}=\underset{x\to 0}{lim}\left[\frac{\frac{x^2}{2!}+\frac{x^3}{3!}+.............}{2}\right]=0$

Therefore,
$\underset{x\to 0}{lim}\frac{e^x-x-1}{2}=0$