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Math Help - Limit without L'Hospital's Rule.

  1. #1
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    Limit without L'Hospital's Rule.

    Hello, I was wondering if there was a nice way to evaluate \displaystyle\smash{\lim_{x\to\ 0}}\dfrac{ln^2{(1+x)}+\ln^2{(1-x)}}{x^2} without using L'Hospital's rule. I'm not clever enough to figure one out. Thank you.
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    Senior Member vincisonfire's Avatar
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    Philosophy rule
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  3. #3
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    It's fairly easy, you just need this little fact: \underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1. Hence,

    \frac{{{\ln }^{2}}(1+x)+{{\ln }^{2}}(1-x)}{{{x}^{2}}}={{\left( \frac{\ln (1+x)}{x} \right)}^{2}}+{{\left( \frac{\ln (1-x)}{x} \right)}^{2}}\to 2, as x\to0.

    (First limit goes to 1, the other one goes to -1; provided the existence of both limits, hence, the result.)
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    It's fairly easy, you just need this little fact: \underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.
    Sorry to be a nag, but could you help me evaluate this without using L'hospitals rule, since this is where I got stuck and had to use it.
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  5. #5
    Senior Member DeMath's Avatar
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    \mathop {\lim }\limits_{x \to 0} \frac{{{{\ln }^2}\left( {1 + x} \right) + {{\ln }^2}\left( {1 - x} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 + x} \right)}^{\frac{1}{x}}}} \right]^2} + \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 - x} \right)}^{\frac{1}{x}}}} \right]^2}.

    \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 + x} \right)}^{\frac{1}{x}}}} \right]^2} = \left\{ \begin{gathered}x = \frac{1}{t}, \hfill \\ x \to 0, \hfill \\t \to \infty  \hfill \\ \end{gathered}  \right\} = \mathop {\lim }\limits_{t \to \infty } {\left[ {\ln {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^2} = {\ln ^2}e = 1.

    \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 - x} \right)}^{\frac{1}{x}}}} \right]^2} = \left\{ \begin{gathered}- x = \frac{1}{t}, \hfill \\x \to 0, \hfill \\t \to \infty  \hfill \\ \end{gathered}  \right\} = \mathop {\lim }\limits_{t \to \infty } {\left[ { - \ln {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^2} = {\left( { - \ln e} \right)^2} = 1.
    Last edited by DeMath; January 11th 2009 at 01:45 PM.
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