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Math Help - lim x->0 [(e^x-1)/x]

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

    Hi, I have been trying to solve this limit for past 4 hours, and I can't figure it out. Can anyone guide me ?

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

    without L'Hosptial, without boundaries
    it should be: 1/2; use lim x->0 [((e^x) -1)/x = 1]


    sorry for wrong topic name, I messed it up, I'm new here.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MarsTeam View Post
    Hi, I have been trying to solve this limit for past 4 hours, and I can't figure it out. Can anyone guide me ?

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

    without L'Hosptial, without boundaries
    it should be: 1/2; use lim x->0 [((e^x) -1)/x = 1]


    sorry for wrong topic name, I messed it up, I'm new here.
    Problem: Compute L=\lim_{x\to0}\left\{\frac{1}{x}-\frac{1}{e^x-1}\right\}

    Solution: Note that \frac{1}{x}-\frac{1}{e^x-1}=\frac{e^x-1-x}{x\left(e^x-1\right)}. Now I am not exactly sure what boundaries are (it may be this) but e^x-1\stackrel{0}{\sim}x and e^x-x-1\stackrel{0}{\sim}1+x+\frac{x^2}{2}-1-x=\frac{x^2}{2} so that L=\lim_{x\to0}\left\{\frac{1}{x}-\frac{1}{e^x-1}\right\}=\lim_{x\to0}\frac{e^x-x-1}{x(e^x-1)}=\lim_{x\to0}\frac{\tfrac{x^2}{2}}{x\cdot x}=\frac{1}{2}.


    Note that if you are weary about using asymptotic equivalences. Merely note that e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}. A particular implications of this is that if |x|<1 we see x+\frac{x^2}\le e^x\-1\le x+\frac{x^2}{2}+\frac{2x^3}{3!} for sufficiently small x.
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