# Evaluate the limit

• Nov 12th 2012, 07:46 AM
calculus123
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.
http://i47.tinypic.com/2r6fwxu.jpg
• Nov 12th 2012, 07:53 AM
Plato
Re: Evaluate the limit
Quote:

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.
http://i47.tinypic.com/2r6fwxu.jpg

Apply L'Hopital's rule twice. What do you get?
• Nov 12th 2012, 08:00 AM
calculus123
Re: Evaluate the limit
I get e^2x*2/1=infinite.
• Nov 12th 2012, 11:24 PM
hollywood
Re: 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