The beauty beauty of L"Hopital"s rule is that it can used multiple times till your indeterminate type goes away.

For instance, friend have

$$\lim_x \to \infty \fracx^2e^x$$

which is an indeterminate type of $\frac\infty\infty$. If you use the rule once you have:

$$\lim_x \to \infty \frac2xe^x$$

However, this is again one indeterminate type of $\frac\infty\infty$. So friend can apply L"Hopital as soon as again until you don"t have an indeterminate form.

$$\lim_x \to \infty \frac2e^x$$

Now, this border is not indeterminate and also it can easily be watched via intuition or graph that the limit must be zero.

Therefore,

$$\lim_x \to \infty \fracx^2e^x = 0$$

Hope that helps!

You are watching: Limit of e^x as x approaches infinity

re-superstructure
cite
follow
answered Feb 21 "16 in ~ 15:48

Jeel ShahJeel shah
$\endgroup$
include a comment |
2
$\begingroup$
Consider $r>0$; then$$\lim_x\to\infty\fracx^re^x=0$$Indeed, you can start indigenous $x0$), for this reason \$\fracx2

Draft saved

name

## Not the price you're spring for? Browse other questions tagged borders or ask your very own question.

0

0

1

2

3

1

1

1

0

stack Exchange Network

thedesigningfairy.comematics ridge Exchange works ideal with JavaScript allowed