limit approximation without using l'hopital

Question

I would like to calculate this limit without using l'hopital:

$$\lim _{x\to \infty }\frac{x^2\cdot \left(1+\dfrac{1}{\ln^2x}\right)^{1+2\ln^2x}-x^2+1}{x-2+x^2}$$

I thought about approximation, but I don't know how.

Answer 1

$$\lim _{x\to \infty }\frac{x^{2}\left(1+\frac{1}{\left(\ln\left(x\right)\right)^{2}}\right)^{\left(1+2\left(\ln\left(x\right)\right)^{2}\right)}-x^{2}+1}{x^{2}+x-2}$$$$=\lim _{x\to \infty }\frac{\color{blue}{\left(1+\frac{1}{\left(\ln\left(x\right)\right)^{2}}\right)^{\left(1+2\left(\ln\left(x\right)\right)^{2}\right)}}-1+\frac{1}{x^{2}}}{1+\frac{1}{x}-\frac{2}{x^{2}}}$$

Then we have:

$$\color{blue}{\lim _{x\to \infty }\left(1+\frac{1}{\left(\ln\left(x\right)\right)^{2}}\right)^{\left(1+2\left(\ln\left(x\right)\right)^{2}\right)}}$$$$=\exp\left(\lim _{x\to \infty }\frac{^{1+2\left(\ln\left(x\right)\right)^{2}}}{\left(\ln\left(x\right)\right)^{2}}\right)$$$$=\exp\left(\lim _{x\to \infty }\frac{1}{\left(\ln\left(x\right)\right)^{2}}+2\right)$$$$=\exp\left(2\right)=e^{2}$$

So finally we have:$$\lim _{x\to \infty }\frac{x^{2}\left(1+\frac{1}{\left(\ln\left(x\right)\right)^{2}}\right)^{\left(1+2\left(\ln\left(x\right)\right)^{2}\right)}-x^{2}+1}{x^{2}+x-2}=e^2-1$$

Answer 2

Dividing by $x^2$ we get $$((1+\frac{1}{ln^2x})^{1+2ln^2x}-1+\frac{1}{x^2})\cdot\frac{1}{\frac{1}{x}-\frac{2}{x^2}+1}=((1+\frac{1}{ln^2x})^{ln^2x})^2\cdot(1+\frac{1}{ln^2x})-1+\frac{1}{x^2})\cdot \frac{1}{\frac{1}{x}-\frac{2}{x^2}+1} $$

And this tends to $e^2-1$ as $x \to \infty$