<s>[INST] solve $\lim_{x \to -\infty} (x +\sqrt{x^2+2x})$. I can't figure out how to mathematically solve $\lim_{x \to -\infty} (x +\sqrt{x^2+2x})$.
According to Desmos, this limit equals $-1$, and this answer was accepted by the software I'm using.
I have tried:
$$\lim_{x \to -\infty} (x +\sqrt{x^2+2x})$$
$$=\lim_{x \to -\infty} (x +\sqrt{x^2+2x}) \cdot \frac{x -\sqrt{x^2+2x}}{x -\sqrt{x^2+2x}}$$
$$=\lim_{x \to -\infty} \frac{-2x}{x - \sqrt{x^2+2x}}$$
$$=\lim_{x \to -\infty} \frac{-2x}{x - x(\sqrt{1+\frac{2}{x}})}$$
But this doesn't seem to point to $-1$ being the limit.
 [/INST] $\lim_{x \rightarrow - \infty} (x + \sqrt{x^2 + 2x})$
$=\lim_{x \rightarrow - \infty} (\frac{-2x}{x - \sqrt{x^2 + 2x}})$
$=\lim_{x \rightarrow - \infty} (\frac{-2}{1 - \sqrt{x^2 + 2x}/x})$
Now use $x = -\sqrt{x^2}$ when $x < 0$ to get:
$=\lim_{x \rightarrow - \infty} (\frac{-2}{1 + {\sqrt{1 + 2/x}}}) = -1$
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