Question

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

• A. no value of n
• B. all values of n
• C. only negative values of n
• D. only positive values of n

Answer 1

Answer: (B)

Case I: $n$ is a positive integer
$\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{n{x}^{n-1}}{e^x}$
$=\underset{x\to \infty }{\lim }\frac{n(n-1)x^{n-2}}{e^x}=\dots =\underset{x\to \infty }{\lim }\frac{n!}{e^x}$
(Using L’Hospital’s Rule repeatedly)
$=0$
Case II: $n$ is a negative integer.
$\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{x^{-m}}{e^x}$
(Putting $n=-m$, where $m$ is a positive integer)
$=\underset{x\to \infty }{\lim }\frac{1}{x^m{e}^x}=\frac{1}{\infty }=0$
Case III: $n=0$
$\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{1}{e^x}=\frac{1}{\infty }=0$
Hence, $\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=0$ for all values of $n$.