Functions Question 182
Question: $ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{n}}{e^{x}}=0 $ for
[IIT 1992]
Options:
A) No value of n
B) n is any whole number
C) $ n=0 $ only
D) $ n=2 $ only
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{n}}{e^{x}}=\underset{x\to \infty }{\mathop{\lim }},n\frac{{x^{n-1}}}{e^{x}}=…… $ $ =\underset{x\to \infty }{\mathop{\lim }},\frac{n!}{e^{x}}=\frac{n!}{\infty }=0 $ ,
where n is any whole number $ (,\because n,! $ is defined for all positive integers including zero).
BETA
