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Math Help - Evaluate the limit

  1. #1
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    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.
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  2. #2
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    Re: Evaluate the limit

    Quote Originally Posted by calculus123 View Post
    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?
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  3. #3
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    Re: Evaluate the limit

    I get e^2x*2/1=infinite.
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  4. #4
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    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.

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