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?
 A: 
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.
A: For negative $x$, we have 
$$\sqrt{1+x^2}=-x\sqrt{1+\frac{1}{x^2}}=-x\left(1+\frac{1}{2x^2}+O(x^{-4})\right)$$
So we have
$$
\lim_{x\rightarrow-\infty}(1+2x^2+2x\sqrt{1+x^2})=\lim_{x\rightarrow-\infty}\left(1+2x^2-2x^2\left(1+\frac{1}{2x^2}+O(x^{-4})\right)\right)=\\
=\lim_{x\rightarrow-\infty}O(x^{-2})=0
$$
You were missing a constant term in your approximation for $\sqrt{1+x^2}$, it can sometimes be dangerous to just reason "by feeling" like you seem to have done, I suggest using Taylor series for rigorous derivations in such cases.
A: 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 $f(x) \to 0$.
A: Hint: We have $$f(x)=x^2\left(\frac{1}{x^2}+2-2\sqrt{1+\frac{1}{x^2}}\right)$$ So our searched limit is $$0$$
A: $$\lim_{x\to-\infty}(1+2x^2+2x\sqrt{1+x^2})$$$$=\lim_{x\to-\infty}(\sqrt{1+x^2}+x)^2$$$$=\lim_{x\to-\infty}\frac{1}{(\sqrt{1+x^2}-x)^2}$$$$=\lim_{x\to\infty}\frac{1}{(\sqrt{1+x^2}+x)^2}$$$$=0$$
