# 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?

#### Solutions Collecting From Web of "Evaluate the $\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$"

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

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

$$\dfrac{\sqrt{1-2h}-1}h=\dfrac{1-2h-1}{h(\sqrt{1-2h}+1)}$$

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}