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Thread: Limit of a natural logarithm

  1. #1
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    Limit of a natural logarithm

    I've some trouble understanding how to work with the natural logarithm when trying to calculate a limit. I just can't seem to fit what I know together:

    - I know that if $\displaystyle x \to 1$, then $\displaystyle \ln x \approx x - 1$, because $\displaystyle \ln(1) = 0$, as $\displaystyle e^0 = 1$. $\displaystyle \ln x, 0 < x < 1$ is a negative, so is $\displaystyle x - 1, x < 1$, so their quotient is a positive.

    - I know the given standard limits: $\displaystyle \lim_{x \to 1}\frac{\ln x}{x -1} = 1$ (following from the above: $\displaystyle \frac{a}{a} = 1$) and $\displaystyle \lim_{x \to 0}\frac{\ln (1 + x)}{x } = 1$

    But, I just can't seem to fit it together when I'm asked to solve an exercise like this: $\displaystyle \lim_{x \to 0}\frac{\ln (1 + x^2)}{x } = x$
    How does this all fit together so I actually get to x? I need this explained in baby steps, it just won't ring a bell at the moment
    Last edited by Lepzed; Jul 19th 2012 at 10:13 PM.
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  2. #2
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    Re: Limit of a natural logarithm

    Okay, here it is. You can do it in any of these ways :-
    1:- $\displaystyle \lim (x->0) \frac{ln(1+x^2)}{x} * \frac{x}{x} $
    So we have multiplied and divided by x
    now using this
    $\displaystyle \lim (f(x)->0) \frac{ln(1+f(x)}{f(x)} = 1 $
    (You can try this yourself. Just apply L'hospital and you will get it)
    we get the limit as 1*x as x->0
    (Where f(x) = x^2)
    so it becomes 0

    2:- Just Apply L'hospital as both the numerator and denominator are tending to zero.
    You will get
    $\displaystyle \frac{2x}{1+x^2} $
    which tends to 0 as x-> 0

    Spare me for my poor Latex. I am trying to learn.
    cheers !
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  3. #3
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    Re: Limit of a natural logarithm

    Made a mistake, brb Thanks so far

    I'm actually not quite sure what's happening here, why do you introduce $\displaystyle \frac{x}{x}$?

    How does this apply to:

    $\displaystyle lim_{x \to 0}\frac{ln(1 + 2x)}{x}$?

    I'm confused to say the least
    Last edited by Lepzed; Jul 20th 2012 at 12:22 AM.
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  4. #4
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    Re: Limit of a natural logarithm

    See the thing inside the log guy is
    $\displaystyle \frac {log(1+x^2)}{x} $
    and to use the result we need $\displaystyle x^2$ in the denominator as well
    so we make it $\displaystyle x^2$, But we need to compensate for that extra x . So we multiply it again.

    Its the same thing that you do in rationalizing ....multiply divide by conjugate. Its a jugglery, that's all.

    $\displaystyle lim_{x \to 0} \frac{1+2x}{x} = Infinity $
    You can see the numerator goes to 1 as denominator goes to 0 so it becomes infinitely huge as you approach 0.
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  5. #5
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    Re: Limit of a natural logarithm

    Sorry i meant ln(1 + 2x), thanks again tho!
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  6. #6
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    Re: Limit of a natural logarithm

    ya so multiply divide by 2 so that denominator becomes 2x and the limit becomes 2 .
    Also you can just apply L'Hospital and get the limit as 2 .
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