<s>[INST] Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this.  I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)
$$
I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick.  That doesn't seem to work.  I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :(
Thanks.
 [/INST] The direct approach of just factoring an $|x|$ from each piece is not fruitful: It leads to
$$ |x| \Big(\sqrt{1 + 2/x} - \sqrt{1 - 2/x}\Big)$$
The first term grows, and the second term tends to $0$, so there's a balance between them.

Multiply top and bottom by the conjugate to find that
\begin{align*}
\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} &= \Big(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}\Big) \left(\frac{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}\right) \\
&= \frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}} \\
&= \frac{-4}{\sqrt{1 + 2/x} + \sqrt{1 - 2/x}}
\end{align*}
since $\sqrt{x^2} = |x| = -x$ for $x < 0$. Can you finish it from here?
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