Question

$\underset{x\to \infty }{lim}\frac{x^n}{e^x}=0$ for

• A. $n=0$ only
• B. $n$ is any whole number
• C. $n=2$ only
• D. no value of $n$

Answer 1

Answer: (A)

$n=0$ only

$\underset{x\to \infty }{lim}\frac{x^n}{e^x}$
$\underset{x\to \infty }{lim}\frac{x^n}{{\sum }_{n=0}^{\infty }\frac{x^n}{n!}}$
$\underset{x\to \infty }{lim}\frac{x^n}{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....\frac{x^n}{n!}+....}$
$\underset{x\to \infty }{lim}\frac{x^n}{x^n(\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{2!x^{n-2}}+\frac{1}{3!x^{n-3}}+....\frac{1}{n!}+....}$
$=n!$
So, ${lim}_{x\to \infty }$ equals 0 when $n=0$