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

Question

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 1

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,
$$=\lim_{x\to \sqrt 2}\left(\frac{(e^{x^2}-e^2)(3x^2-6\sqrt 2 x+5)+(x^3-3\sqrt 2x^2+5x)(2xe^{x^2})}{\frac{(2x-3\sqrt 2)}{x^2-3\sqrt 2 x+5}\cdot(2x-3\sqrt 2)+\ln(x^2-3\sqrt 2 x+5)(2)}\right)$$ $$=\left(\frac{0+(\sqrt 2)(2\sqrt 2e^2)}{\frac{(-\sqrt 2)}{1}\cdot(-\sqrt 2)+0}\right)=\color{red}{2e^2}$$

Answer 2

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

$$=e^2\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{x^2-3\sqrt2 x+4}\cdot\dfrac1{\left(\lim _{x\to \sqrt{2}}\dfrac{\ln(1+x^2-3\sqrt2 x+4)}{x^2-3\sqrt2 x+4}\right)^2}$$

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 $$e^2\cdot4\cdot\dfrac12$$

Answer 3

$$ \begin{aligned} \lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2\left(1-x^2\right)}{\left(ln\left(x^2-3\sqrt{2}x+5\right)\right)^2}\right) & = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(1-\left(t+\sqrt{2}\right)^2\right)}{\left(ln\left(\left(t+\sqrt{2}\right)^2-3\sqrt{2}\left(t+\sqrt{2}\right)+5\right)\right)^2}\right) \\& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(-t^2-2\sqrt{2}t-1\right)}{\ln \:^2\left(t^2-\sqrt{2}t+1\right)}\right) \\& = \lim _{t\to 0}\left(\frac{e^2+2\sqrt{2}e^2t+5e^2t^2+o\left(t^2\right)+e^2\left(-t^2-2\sqrt{2}t-1\right)}{2t^2+o\left(t^2\right)}\right) \\& = \color{red}{2e^2} \end{aligned} $$ Solved with substitution $t = x-\sqrt2$ and Taylor expansion