calculus - How to solve $lim _{xto sqrt{2}}left(frac{e^{x^2}+e^2(1-x^2)}{[ln(x^2-3sqrt{2}x+5)]^2}right)$?

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
$$\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[ln(x^2-3\sqrt{2}x+5)]^2}\right)$$

Answer

$$
\begin{aligned}
\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2\left(1-x^2\right)}{\left(ln\left(x^2-3\sqrt{2}x+5\right)\right)^2}\right)
& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(1-\left(t+\sqrt{2}\right)^2\right)}{\left(ln\left(\left(t+\sqrt{2}\right)^2-3\sqrt{2}\left(t+\sqrt{2}\right)+5\right)\right)^2}\right)
\\& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(-t^2-2\sqrt{2}t-1\right)}{\ln \:^2\left(t^2-\sqrt{2}t+1\right)}\right)
\\& = \lim _{t\to 0}\left(\frac{e^2+2\sqrt{2}e^2t+5e^2t^2+o\left(t^2\right)+e^2\left(-t^2-2\sqrt{2}t-1\right)}{2t^2+o\left(t^2\right)}\right)
\\& = \color{red}{2e^2}
\end{aligned}
$$

Solved with substitution $t = x-\sqrt2$ and Taylor expansion