Hints about the limit $\lim_{x \to \infty} ((1+x^2)/(x+x^2))^{2x}$ without l'Hôpital's rule?

Question

I've tried to evaluate $\lim_{x \to \infty} \left(\frac {1+x^2}{x+x^2}\right)^{2x}$ as $$\lim_{x \to \infty} \left(\left(\frac {1+ \frac{1}{x^2}}{1+ \frac{1}{x}}\right)^{x}\right)^{2}$$ So the denominator goes to $e^2$, but I don't know how to solve the numerator, because of the $x^2$. Any hint?

Thanks in advance!

Answer 1

HINT

We have

$$\left(\frac {1+x^2}{x+x^2}\right)^{2x}=\left(\frac {x+x^2+1-x}{x+x^2}\right)^{2x}=\left[\left(1+\frac {1-x}{x+x^2}\right)^{\frac {x+x^2}{1-x}}\right]^{\frac {2x(1-x)}{x+x^2}}$$

Answer 2

hint

use the equality

$$\frac{1+x^2}{x+x^2}=1-\frac{x-1}{x+x^2}$$ $$=1+X$$

the function can be written as

$$e^{2x\ln(1+X)}=e^{2xX\frac{\ln(1+X)}{X}}$$

and observe that when $x\to+\infty, \; X\to 0$,

$$\frac{\ln(1+X)}{X}\to 1$$ and $$xX\to -1.$$

You should find $e^{-2}$.

Answer 3

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

So $\lim\limits_{x \to \infty}\left(\frac {1+x^2}{x+x^2}\right)^{2x}=e^{-2}$