<s>[INST] 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
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
 [/INST] Notice, one can easily apply L'Hospital's rule for $\frac 00$  form as follows $$\lim_{x\to \sqrt 2}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt 2 x+5)]^2}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{2xe^{x^2}-2xe^2}{\frac{2\ln(x^2-3\sqrt 2 x+5)}{x^2-3\sqrt 2 x+5}\cdot (2x-3\sqrt 2)}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{x(e^{x^2}-e^2)(x^2-3\sqrt 2 x+5)}{(2x-3\sqrt 2)\ln(x^2-3\sqrt 2 x+5)}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{(e^{x^2}-e^2)(x^3-3\sqrt 2 x^2+5x)}{(2x-3\sqrt 2)\ln(x^2-3\sqrt 2 x+5)}\right)$$
applying L'Hospital's rule for $\frac 00$ form,
$$=\lim_{x\to \sqrt 2}\left(\frac{(e^{x^2}-e^2)(3x^2-6\sqrt 2 x+5)+(x^3-3\sqrt 2x^2+5x)(2xe^{x^2})}{\frac{(2x-3\sqrt 2)}{x^2-3\sqrt 2 x+5}\cdot(2x-3\sqrt 2)+\ln(x^2-3\sqrt 2 x+5)(2)}\right)$$
$$=\left(\frac{0+(\sqrt 2)(2\sqrt 2e^2)}{\frac{(-\sqrt 2)}{1}\cdot(-\sqrt 2)+0}\right)=\color{red}{2e^2}$$
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