Question

If $n=2$ then the value of $\underset{x\to 0}{lim}\frac{e^x-1-x-\frac{x^2}{2!}-....-\frac{x^n}{n!}}{x^{n+1}}$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{6}$

$\underset{x\to 0}{lim}\frac{e^x-1-x-\frac{x^2}{2!}-......-\frac{x^n}{n!}}{x^{n+1}}$
Using expansions, we have:
$\underset{x\to 0}{lim}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}...)-1-x-\frac{x^2}{2!}}{x^3}$

$=\frac{1}{3!}=\frac{1}{6}$