1. Evaluate 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.

2. Re: Evaluate the limit

Originally Posted by calculus123
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.
Apply L'Hopital's rule twice. What do you get?

3. Re: Evaluate the limit

I get e^2x*2/1=infinite.

4. Re: Evaluate the limit

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

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

- Hollywood