Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\infty}(1+2x^2-2x^2)=1$$
However, if you plot the function in Desmos or you do it with a calculator, you will find that $f(x\rightarrow-\infty)=0$
What am I missing?

$$\lim_{x\to-\infty}(1+2x^2+2x\sqrt{1+x^2})$$$$=\lim_{x\to-\infty}(\sqrt{1+x^2}+x)^2$$$$=\lim_{x\to-\infty}\frac{1}{(\sqrt{1+x^2}-x)^2}$$$$=\lim_{x\to\infty}\frac{1}{(\sqrt{1+x^2}+x)^2}$$$$=0$$
