How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$

$$\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
$$=e^2\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{x^2-3\sqrt2 x+4}\cdot\dfrac1{\left(\lim _{x\to \sqrt{2}}\dfrac{\ln(1+x^2-3\sqrt2 x+4)}{x^2-3\sqrt2 x+4}\right)^2}$$
Now,
$$\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{(x^2-3\sqrt2 x+4)^2}=\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{(x^2-2)^2}\cdot\lim _{x\to \sqrt2}\dfrac{(x+\sqrt2)^2}{(x-2\sqrt2)^2}$$
$$=(-2)^2\lim_{y\to0}\dfrac{e^y-1-y}{y^2}$$
Finally use How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?
to get the limit to be $$e^2\cdot4\cdot\dfrac12$$
