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

The correct option is B all values of $n$

Case I. $n$ is a positive integer.

$\underset{x\to \infty }{lim}\frac{x^2}{e^x}=\underset{x\to \infty }{lim}\frac{{nx}^{n-1}}{e^x}$

$=\underset{x\to \infty }{lim}\frac{n(n-1)x^{n-2}}{e^x}=...=\underset{x\to \infty }{lim}\frac{n!}{e^x}$

[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 repeatedly]

$=\underset{x\to \infty }{lim}\frac{1}{x^m{e}^x}=\frac{1}{\infty }=0$

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 $n.$