<s>[INST] How to compute $\lim\limits_{x \to 0} \left(\frac{e^{x^2} -1}{x^2}\right)^\frac{1}{x^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 0} \left(\frac{e^{x^2} -1}{x^2}\right)^\frac{1}{x^2}$$

Note: In a previous version of this question the limit was written as $\left(\frac{(e^{x})^2 -1}{x^2}\right)^\frac{1}{x^2}$.

 [/INST] Let 
$$y = \left(\frac{e^{2x} -1}{x^2}\right)^\frac{1}{x^2}.$$
Then
$$\ln y =  \frac{1}{x^2} \ln\left(\frac{e^{2x} -1}{x^2}\right).$$
Notice that 
$$\lim_ \limits{x \to 0^+} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^+} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)=\infty,$$
but 
$$\lim_ \limits{x \to 0^-} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^-} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)$$
is undefined since you have $-\infty$ inside the $\log$.
Therefore the limit does not exist
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