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!

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answered Feb 21 "16 in ~ 15:48

Jeel ShahJeel shah

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Consider $r>0$; then$$\lim_x\to\infty\fracx^re^x=0$$Indeed, you can start indigenous $x0$), for this reason $\fracx2

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