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

1. ## 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.

2. Philosophy rule

3. 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.)

4. Originally Posted by Krizalid
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.

5. $\mathop {\lim }\limits_{x \to 0} \frac{{{{\ln }^2}\left( {1 + x} \right) + {{\ln }^2}\left( {1 - x} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 + x} \right)}^{\frac{1}{x}}}} \right]^2} + \mathop {\lim }\limits_{x \to 0} {\left[ {\ln {{\left( {1 - x} \right)}^{\frac{1}{x}}}} \right]^2}.$

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

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