Question

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