Could someone help me solve this using L'Hopital rule? I end up getting infinity/infinity which equals infinity but that's not the right asnwer.

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- Nov 12th 2012, 07:46 AMcalculus123Evaluate the limit
Could someone help me solve this using L'Hopital rule? I end up getting infinity/infinity which equals infinity but that's not the right asnwer.

http://i47.tinypic.com/2r6fwxu.jpg - Nov 12th 2012, 07:53 AMPlatoRe: Evaluate the limit
- Nov 12th 2012, 08:00 AMcalculus123Re: Evaluate the limit
I get e^2x*2/1=infinite.

- Nov 12th 2012, 11:24 PMhollywoodRe: Evaluate the limit
That's correct. The limit of denominator and numerator are both infinity, so you have $\displaystyle \frac{\infty}{\infty}$ and can apply L'Hopital's rule. Taking the derivative of numerator and denominator gives $\displaystyle \frac{2e^{2x}}{2x}=\frac{e^{2x}}{x}$ and again you have $\displaystyle \frac{\infty}{\infty}$. So apply L'Hopital's rule again to get $\displaystyle \frac{2e^{2x}}{1}=2e^{2x}$. Since this goes to infinity as x goes to infinity, you can conclude that $\displaystyle \lim_{x\rightarrow{\infty}}\frac{e^{2x}}{x^2}=$$\displaystyle \infty$.

You probably didn't mean to say this in your first post, but I should point out that you can't assume that $\displaystyle \frac{\infty}{\infty}$ is infinity - it might be, but it could also be any positive number or zero.

- Hollywood