Calculate limit using L'Hopital Rule -> $ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} $ I have been trying to solve the following problem, but I seem t have some difficulties.
$$ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} $$
I tryied to solve it, and the answer I got didn't seem to be right, so I tryed to plot it to have an ideia, and I reached the conclusion that it was wrong.
This is what I did on my second attempt:
$$ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} = \lim_{x \to 0} \left ( \frac{x^2 + 1}{x^2} \right )^{x^2} = \left [ \left ( \frac{(0)^2 + 1}{(0)^2} \right )^{(0)^2} \right ] = \left [ \infty ^0 \right ] \Rightarrow $$
$$ \Rightarrow \lim_{x \to 0} e^{x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right )} = e^{ \lim_{x \to 0} x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right )} $$
Then, to make the calculations easier:
$$ \lim_{x \to 0} x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right ) = \left [ (0)^2 \ln \left ( \frac{(0)^2 + 1}{(0)^2} \right ) \right ] = \left [ 0 \cdot \ln \left ( \frac{ 1}{0} \right ) \right ] $$
How do I proced now? I mean, How can I solve $ \left [ 0 \cdot \ln \left ( \frac{ 1}{0} \right ) \right ] $? There is no signal on the 0. If it was $ 0^{+} $, I would know it would aproximate to $ -\infty $.
Can you help me please?
I do not intend to have the solution to the limit, but some aid to solve this problem.
Thanks in advance, Saclyr.
 A: HINT: Substitute $n=x^2$ then it becomes to$$\Big(1+\dfrac{1}{n}\Big)^n=\dfrac{(n+1)^n}{n^n}.$$ Also $x\to 0$ is equivalent to $n\to 0^+.$ 
A: $$\left(1+\frac{1}{x^2}\right)^{x^2}=\exp\left(x^2\ln\left(1+\frac{1}{x^2}\right)\right)$$
$$x^2\ln\left(1+\frac{1}{x^2}\right)=\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}$$
therefore
$$\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)=\lim_{x\to 0}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}$$
$$\lim_{x\to 0^+}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}=\lim_{u\to\infty }\frac{\ln(1+u^2)}{u^2}\underset{Hop.}{=}\lim_{u\to\infty }\frac{2u}{(1+u^2)2u}=\lim_{u\to\infty }\frac{1}{1+u^2}=0$$
and
$$\lim_{x\to 0^-}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}=\lim_{u\to-\infty }\frac{\ln(1+u^2)}{u^2}\underset{Hop.}{=}\lim_{u\to-\infty }\frac{2u}{(1+u^2)2u}=\lim_{u\to-\infty }\frac{1}{1+u^2}=0$$
and thus,
$$\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)=0.$$
Moreover, $x\mapsto e^x$ is continuous in $x=0$, therefore
$$\lim_{x\to 0}\exp\left(x^2\ln\left(1+\frac{1}{x^2}\right)\right)=\exp\left(\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)\right)=e^0=1.$$
A: rewrite it in the form $$e^{\lim_{x \to 0}\frac{\ln\left(\frac{x^2+1}{x^2}\right)}{\frac{1}{x^2}}}$$
