How to Solve This Exponential Limit without Derivate / L'Hôspital's Rule

can someone teach me how can I solve this limit without using the L'Hopital's Rule?

$$\lim_{x\to 0} \left( \frac{2+x^{2}}{2-x^{2}} \right)^{\frac{1}{x^2}}$$

Thanks in advance.

Answer

$$\frac{2+x^2}{2-x^2}=1+\frac{2x^2}{2-x^2}=1+\frac{x^2}{1-\frac{x^2}{2}}$$
and $y=\frac{x^2}{1-\frac{x^2}{2}}\to 0$, as $x\to 0$. Hence
$$\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}=(1+y)^{1/y}\to e,$$

as $x\to 0$.

Finally

$$\left(\frac{2+x^2}{2-x^2}\right)^{1/x^2}=\left(\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}\right)^{\frac{1}{1-\frac{x^2}{2}}}\to e,$$
since, if $f(x)\to e$ and $g(x)\to 1$, then $f(x)^{g(x)}\to e$.