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?

Multiply top and bottom by a conjugate:
$$\begin{align*}f(x) & = \dfrac{1+2x^2+2x\sqrt{1+x^2}}{1}\cdot \dfrac{1+2x^2-2x\sqrt{1+x^2}}{1+2x^2-2x\sqrt{1+x^2}} \\ & = \dfrac{1}{1+2x^2-2x\sqrt{1+x^2}}\end{align*}$$
Now, as $x \to -\infty$, $1+2x^2-2x\sqrt{1+x^2} \to \infty$, so $f(x) \to 0$.
