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Math Help - limits

  1. #1
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    limits

    can somebody please explain this problem to me:


    lim [ (x(e^(-2x+1))) / ((x^2)+x)]
    x-->0



    i get 0/0 = undefined
    but my teacher tells me that it is not correct.

    please help
    thank you
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  2. #2
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    Nevermind.
    Last edited by Defunkt; October 9th 2009 at 06:49 AM.
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  3. #3
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    Quote Originally Posted by rahul14 View Post
    lim [ (x(e^(-2x+1))) / ((x^2)+x)]
    x-->0
    \lim _{x \to 0} \frac{xe^{-2x+1}}{x^2+x}=\lim _{x \to 0}\frac{e^{-2x+1}}{x+1}\to e
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  4. #4
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    Quote Originally Posted by rahul14 View Post
    can somebody please explain this problem to me:


    lim [ (x(e^(-2x+1))) / ((x^2)+x)]
    x-->0



    i get 0/0 = undefined
    but my teacher tells me that it is not correct.

    please help
    thank you
    do you mean
     \lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}

    if yes , then
     \lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}=\lim_{x \to 0} \frac {x.e^{-2x+1}}{x(x+1)}

    = \lim_{x \to 0} \frac {e^{-2x+1}}{x+1}= \frac {e^{0+1}}{0+1} =e
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    Yes, \frac{0}{0} is correct. Have you learned l'H˘pital's rule?
    No, 0/0 is NOT correct for the limit! It's not even a number. And I surely would not recommend "L'Hopital's rule" for something as simple as this.

    \frac{xe^{-2x+1}}{x^2+ x}= \frac{xe^{-2x+1}}{x(x+1)}
    which is equal to \frac{e^{-2x+1}}{x+1} as long as x is not 0. They have the same limit so just put x= 0 in \frac{e^{-2x+1}}{x+1}
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