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L' Hospital's Rule Application

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data
    [tex] \lim_{x\rightarrow 0} \frac{xe^2x+xe^x-2e^2x+2e^x}{(e^x-1)^3}[/tex]



    2. Relevant equations
    L' Hospital's Rule
    [tex]\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}= \lim_{x\rightarrow\0} \frac{f'(x)}{g'(x)}[/tex]



    3. The attempt at a solution

    Phew, I think my hold on the latex codes are solid now, thanks Status. Okay, here's what I've worked so far, and where I'm stuck at.

    Just as Hosp. Rule says, I found the derivative of the numerator and denominator functions seperately by using the chain rule:

    [tex]\lim_{x\rightarrow 0} \frac{e^2x(2x+1)+e^x(x+1)-4e^2x+2e^x}{(3e^x)(e^x-1)^2}[/tex]

    I noticed that the function was still indeterminate due to the denominator going to 0, so I found the derivatives of the numerator and denominator again;

    But my result showed that the denominator went to 0 again because its dervivative is:

    [tex] 3e^x(e^x-1)(3e^x-1)[/tex]

    I must be doing something wrong.
     
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2

    StatusX

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    Homework Helper

    You'll eventually get there, but it'll take a few times going through the process. You can make the algebra alot easier if you make the substitution u=e^x (so that the limit is now as u->1).
     
  4. Dec 10, 2006 #3
    I did it quickly, so my answer might be off, but keep doing L'Hopital's rule until you get a proper answer.

    (I got [tex]\frac{1}{6}[/tex], by the way.

    Basically, just keep differentiating until you get rid of the 1 in the [tex]e^{x}-1[/tex] bracket, as then you'll get a proper answer.
     
  5. Dec 10, 2006 #4
    Jeez again, man our teacher is really making us grind out these calculations...well okay thank you fellas.
     
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