calculus - Evaluate the $lim_{x to -infty} (x + sqrt{x^2 + 2x})$

Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$

I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.

This is what I've done so far

\begin{align}
\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x}) &= \lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})\left(\frac{x-\sqrt{x^2 + 2x}}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{x^2 - (x^2 + 2x)}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{-2x}{x-\sqrt{x^2 + 2x}}\right)\\
\\
\end{align}

And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.

Plugging it into WolframAlpha shows that the correct answer is $-1$

Any suggestions on what to do next?

Answer

$$\lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) =\lim _{ x\rightarrow -\infty }{ \left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) =\lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x } } } = } } \\ =\lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x+x\sqrt { 1+\frac { 2 }{ x } } } = } \lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x } } \right) } = } -1$$