MCQ
$\lim _{x \rightarrow \infty} \frac{x^n}{ e ^x}=0$ for
  • A
    no value of $n$
  • n is any whole number
  • C
    $n =0$ only
  • D
    $n =2$ only

Answer

Correct option: B.
n is any whole number
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow \infty} \frac{x^{ n }}{ e ^x}=\lim _{x \rightarrow \infty} \frac{ n x^{ n -1}}{ e ^x}$
$=\lim _{x \rightarrow \infty} \frac{ n ( n -1) x^{ n -2}}{ e ^x}=\lim _{x \rightarrow \infty} \frac{ n !}{ e ^x}=\frac{ n !}{\infty}=0$,
where n is any whole number... $[\because n !$ is defined for all positive integers including zero]

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