<s>[INST] 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?
 [/INST] $$\lim _{ x\to -\infty  } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x }  }  \right) =\lim _{ x\rightarrow -\infty  }{ \left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x }  \right)  }  }  \right) =\lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x }  }  } = }  } \\ =\lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x+x\sqrt { 1+\frac { 2 }{ x }  }  } = } \lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x }  }  \right)  } = } -1$$
</s>