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Math Help - Limit Help

  1. #1
    Junior Member
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    Limit Help

    How do you solve these limits?

    \lim_{x\to0} \ln\frac{e^x-1}{x}

    and

    \lim_{x\to0} \frac{\sin(3x)}{5x}
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Mike9182 View Post
    How do you solve these limits?

    \lim_{x\to0} \ln\frac{e^x-1}{x}

    and

    \lim_{x\to0} \frac{\sin(3x)}{5x}
    Usually one would solve them using L'Hospital's Rule like this:

    \lim_{x\rightarrow 0}\frac{\mathrm{e}^x-1}{x}=\lim_{x\to 0}\frac{\mathrm{e}^x}{1}=\frac{\mathrm{e}^0}{1}=1

    EDIT: I have simply dropped the \ln so let me do it (hopefully) right
    \lim_{x\to0} \ln\frac{e^x-1}{x}=\ln \lim_{x\to0} \frac{e^x-1}{x}=\ln \lim_{x\to 0}\frac{\mathrm{e}^x}{1}=\ln\frac{\mathrm{e}^0}{1}  =\ln 1=0
    END OF EDIT

    and

    \lim_{x\to 0}\frac{\sin(3x)}{5x}=\lim_{x\to 0}\frac{\cos(3x)\cdot 3}{5}=\frac{\cos(3\cdot 0)\cdot 3}{5}=\frac{3}{5}
    Last edited by Failure; August 7th 2009 at 11:43 AM.
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  3. #3
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    ok, I must be supposed to graph them then because they showed up in chapter 2 of my book which is the same chapter limits were introduced in.
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