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)

all values of $n$

Case I. $n$ is a positive integer

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

$\underset{x\to \infty }{lim}\frac{e^n}{e^x}=\underset{x\to \infty }{lim}\frac{x^{-m}}{e^x}$

$[$Putting $n=-m,$ where $m$ is a positive integer$]$

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

Case III. $n=0$

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

Hence, $\underset{x\to \infty }{lim}\frac{x^n}{e^x}=0$ for all values of $n.$