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Math Help - Simplifying logs

  1. #1
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    Simplifying logs

    see this pic: http://i56.tinypic.com/2j5fgwx.jpg

    Thanks!
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  2. #2
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    The limit of logs is log of limits.
     \displaystyle \lim _{x \to a} \ln (f(x)) = \ln (\lim _{x \to a} f(x))
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  3. #3
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    but mine is different....there is simply no ln in step 2...
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  4. #4
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    Quote Originally Posted by janedoe View Post
    but mine is different....there is simply no ln in step 2...
    But the limit in step 2 determines the limit in step one.
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  5. #5
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    Hello, janedoe!

    I'm sure it was a typo . . .


    It should have said:

    \displaystyle \lim_{x\to\infty}\bigg[x - \ln(x^2+1)\bigg] \;=\;\lim_{x\to\infty}\bigg[\ln(e^x) - \ln(x^2+1)\bigg]

    . . \displaystyle =\;\lim_{x\to\infty}\bigg[\ln\left(\frac{e^x}{x^2+1}\right)\bigg] \;=\;\ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{x^2+1}\right)\bi  gg]


    Apply L'Hopital: . \displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2x}\right)\bigg]


    Apply L'Hopital: . \displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2}\right)\bigg] \;=\;\ln(\infty) \;=\;\infty


    Therefore: . \displaystyle \lim_{x\to\infty}\bigg[x - \ln(x^2+1)\bigg] \;=\;\infty

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