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Math Help - solving a ln limit algebraically

  1. #1
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    solving a ln limit algebraically

    The limit as x->0^+ (from above zero) of ln(x+1)/x is 1, I know this from punching numbers into my calculator (and from Maple). My question is how can i show this algebraically?
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  2. #2
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    Re: solving a ln limit algebraically

    Hey kingsolomonsgrave.

    For this problem you get a 0/0 case which is where L'Hopitals rule comes in (and since you are approaching it from the right and not the left you can use this: if you approached it from the left then it wouldn't exist altogether in the real numbers).

    d/dx[ln(x+1)] = 1/(x+1) and d/dx(x) = 1. This will give a limit that can be evaluated.

    If you are not familiar with L'Hopitals rule it is basically a rule that you can use if you get an indeterminate form like 0/0 or infinity/infinity type limits.
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  3. #3
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    Re: solving a ln limit algebraically

    Hello, kingsolomonsgrave!

    Do you know this definition of e?

    . . \lim_{u\to\infty}\left(1 + \tfrac{1}{u}\right)^u \;=\;e



    \text{Show that: }\:\lim_{x\to0^+}\,\frac{\ln(x+1)}{x} \:=\:1

    We have: . \lim_{x\to0^+}\tfrac{1}{x}\ln(x+1) \;=\;\lim_{x\to0}\ln(1+x)^{\frac{1}{x}}

    Let u =\tfrac{1}{x} \quad\Rightarrow\quad x \,=\,\tfrac{1}{u}
    Note that: if x\to 0, then u \to \infty.

    \text{So we have: }\:\lim_{u\to\infty}\left[\ln\left(1 + \tfrac{1}{u}\right)^u\right] \;=\;\ln\underbrace{\left[\lim_{u\to\infty}\left(1 + \tfrac{1}{u}\right)^u\right]}_{\text{This is }e} \;=\;\ln(e) \;=\;1
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  4. #4
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    Re: solving a ln limit algebraically

    Quote Originally Posted by kingsolomonsgrave View Post
    The limit as x->0^+ (from above zero) of ln(x+1)/x is 1, My question is how can i show this algebraically?
    Napier's inequality: 0<a<b implies that \frac{1}{b}<\frac{\ln(b)-\ln(a)}{b-a}<\frac{1}{a}.

    Use that to get \frac{1}{x+1}<\frac{\ln(x+1)}{x}<1.
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  5. #5
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    Re: solving a ln limit algebraically

    thanks so much!
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