logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

$\;\lim \limits_{x \to \infty}\large\frac{x^{2}}{e^{x}}$ is

\[\begin{array}{1 1}(1)2&(2)0\\(3)\infty&(4)1\end{array}\]

Can you answer this question?
 
 

1 Answer

0 votes
$\lim \limits_{x \to \infty } \large\frac{x^2}{e^x} =\frac{\infty^2}{e^x}=\frac{\infty}{\infty}$
By L Hospital Rule
$\lim \limits_{x \to \infty } \large\frac{x^2}{e^x} $$=\lim \limits_{x \to \infty } \large\frac{2x}{e^x}$
$\qquad=\lim \limits_{x \to \infty } \large\frac{x^2}{e^x}$
$\qquad= \large\frac{2}{e^x}=\frac{2}{\infty}=$$0$
Hence 2 is the correct answer.
answered May 19, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...