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?
 A: $$\lim _{ x\to -\infty  } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x }  }  \right) =\lim _{ x\rightarrow -\infty  }{ \left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x }  \right)  }  }  \right) =\lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x }  }  } = }  } \\ =\lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x+x\sqrt { 1+\frac { 2 }{ x }  }  } = } \lim _{ x\rightarrow -\infty  }{ \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x }  }  \right)  } = } -1$$
A: HINT:
Set $-1/x=h\implies h\to0^+, h>0$
$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=+\dfrac{\sqrt{1-2h}}h$
Now rationalize the numerator to get
$$\dfrac{\sqrt{1-2h}-1}h=\dfrac{1-2h-1}{h(\sqrt{1-2h}+1)}$$
A: This is based on @Battani's answer but with a more in-depth explanation 
\begin{align}
            \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) \\ \\
&\text{Now because $\sqrt{x^2}$ = $|x|$}  \\ \\
               &=  \lim _{ x\rightarrow -\infty  } \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x }  }  } \\ \\
\text{Recall that } & \ |x| = \begin{cases}
x &\text{ if } \ \ x \geq 0\\
- x &\text{ if } \ \ x < 0\\
\end{cases}  
 \\  \\
&\text{$x$ is approaching $-\infty$} \\ \\ 
&\therefore \ \ \ \  |x| = -x 
\\ \\
         &=   \lim _{ x\rightarrow -\infty  } \frac { -2x }{ x- (-x)\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)  }    \\ \\
&= \lim _{ x\rightarrow -\infty  } \frac { -2 }{ 1+\sqrt { 1+\frac { 2 }{ x }  }    } \\\\
&= -1
\end{align}
A: This solution is probably flawed, but I will post this anyway.
Note that 
$$\lim_{x \to \ -\infty} \left(x + \sqrt{x^2 + 2x}\right)$$
$$=\lim_{x \to \ -\infty} \left(x + \sqrt{(x+1)^2-1}\right).$$
The $-1$ in the radical is basically negligible compared to the $(x+1)^2$, so let us ignore that. Doing this gives 
$$\lim_{x \to \ -\infty} \left(x + \sqrt{(x+1)^2}\right)$$
$$=\lim_{x \to \ -\infty} \left(x \pm (x+1)\right)$$
$-(x+1)$ is positive, while $x+1$ is negative for sufficiently low $x$. Square roots are always nonegative for real $x$, so we should add $-(x+1)$. Doing so gives
$$\lim_{x \to \ -\infty} \left(x - (x+1)\right)$$
$$=\lim_{x \to \ -\infty} -1$$
$$=\boxed{-1}.$$
