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Homework Help: Where am i wrong in this limits question?

  1. Jul 22, 2011 #1
    question is
    limx->0 (1+x+x^2-e^x)/x^2
    How i did it.
    =>limx->0 (1-e^x)/x^2+(x+x^2)/x^2
    =>limx->0 ((1-e^x)/x)*1/x+1/x+1
    using limx->0(e^x-1)/x=1
    =>limx->0 (-1)*1/x+1/x+1
    =>limx->0 1
    =>Answer=1. (hope you understand my writing way. Where are list of tags for this forum)
    I think i do nothing wrong but this is wrong answer. Applying L'Hospital rule it's answer will come out to be 1/2 which is correct answer. Tell me where i am wrong
     
    Last edited: Jul 22, 2011
  2. jcsd
  3. Jul 22, 2011 #2
    =>limx->0 ((1-e^x)/x)*1/x+1/x+1
    using limx->0(e^x-1)/x=1
    =>limx->0 (-1)*1/x+1/x+1
    Here you are wrong
     
  4. Jul 22, 2011 #3
    dear here i change (1-e^x)/x by 1 which is an identity. I use this identity also that lim f(x).g(x)=lim f(x).lim g(x) and lim [f(x)+g(x)]=lim f(x)+lim g(x)
    friend try to explain it in better way.
     
  5. Jul 22, 2011 #4

    Ray Vickson

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    Your problem is that you look only at the "first-order" terms in (1 - e^x), so as to conclude that lim{ [(1 - e^x)/x] *(1/x) + 1/x] } = 0 as x --> 0. This is false. When we look at (1-e^x) we need to keep also terms of second order in small x, because those will give the nonzero part of lim { [(1 - e^x)/x] *(1/x) + 1/x] }. In other words: if e^x = 1 + x + r(x), we have (1 - e^x)/x = (-x - r(x) )/x = -1 - r(x)/x, hence (1/x)[(1-e^x)/x] = -1/x - r(x)/x^2, so lim { [(1 - e^x)/x] *(1/x) + 1/x] } = - lim r(x)/x^2. This last limit is not zero.

    RGV
     
  6. Jul 22, 2011 #5
    (1-e^x)/x =1 is not an identity. an identity is an algebraic equation that is true for any x belonging to a certain domain. like (1+x)^2 = x^2 + 2x + 1 which is valid for any x in real numbers. but (1-e^x)/x = 1 is only true when x approaches zero. It's not an identity. as Ray Vickson said, you need to consider more terms of the e^x series.
     
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