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?


We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$,

Not exactly. Both $\sqrt{1+x^2}$ and $|x|$ tend to $\infty$ as $x \to -\infty$.  I think what you mean that each is asymptotic to the other, because their ratio tends to $1$ as $x\to-\infty$.  But that doesn't mean you can replace one with the other, as we can see here.
As $x\to -\infty$, this limit takes the form $\infty - \infty$, which is an indeterminate form.  One way to evaluate such a limit is to rewrite it in a form which is not indeterminate.  For convenience, let's first write it as
$$
\lim_{x\to\infty} \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right)
$$
Here we are just substituting $-x$ for $x$; it doesn't change the sign of $x^2$ or $\sqrt{1+x^2}$, but it does change the sign of $x$.  The typical algebraic trick for dealing with radicals such as this is to multiply and divide by the conjugate radical:
\begin{align*}
    1 + 2x^2 - 2 x\sqrt{1+x^2}
    &= \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right) \frac{1 + 2x^2 + 2 x\sqrt{1+x^2}}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
    &= \frac{\left((1 + 2x^2) - 2 x\sqrt{1+x^2}\right)\left((1 + 2x^2) + 2 x\sqrt{1+x^2}\right)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
    &= \frac{(1+2x^2)^2 - \left(2x \sqrt{1+x^2}\right)^2}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
    &= \frac{(1+4x^2+4x^4) - 4x^2(1+x^2)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
    &= \frac{1}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
\end{align*}
Now as $x\to\infty$, the denominator tends to $\infty$ while the numerator is a constant $1$.  Therefore the quotient, and the original expression, tends to zero.
