Find the limit of $x +\sqrt{x^2 + 8x}$ as $x\to-\infty$ $$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}$$
I multiplied it by the conjugate:
$\frac{-8x}{x - \sqrt{{x^2} + 8x}}$
I can simplify further and get:
$\frac{-8}{1-\sqrt{1+\frac{8}{x}}}$
I think there is an error with my math, because the denominator should probably be a 2.
I'm stuck on this one. I graphed it so I know the limit is -4, but I can't calculate it. Thanks a lot for the help!
 A: Notice, $$\lim_{x\to -\infty}(x+\sqrt{x^2+8x})$$
substituting $x=-x$
$$=\lim_{x\to \infty}(-x+\sqrt{x^2-8x})$$
$$=\lim_{x\to \infty}(\sqrt{x^2-8x}-x)\frac{(\sqrt{x^2-8x}+x)}{(\sqrt{x^2-8x}+x)}$$
$$=\lim_{x\to \infty}\frac{x^2-8x-x^2}{\sqrt{x^2-8x}+x}$$
$$=-8\lim_{x\to \infty}\frac{x}{x\sqrt{1-\frac{8}{x}}+x}$$
$$=-8\lim_{x\to \infty}\frac{x}{x\left(\sqrt{1-\frac{8}{x}}+1\right)}$$
$$=-8\lim_{x\to \infty}\frac{1}{\sqrt{1-\frac{8}{x}}+1}$$
$$=-8\left(\frac{1}{\sqrt{1-0}+1}\right)$$
$$=-\frac{8}{2}=\color{red}{-4}$$
A: We have
\begin{align}
\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}&=\lim_{x\to -\infty}\frac{(x +\sqrt{x^2 + 8x})(x -\sqrt{x^2 + 8x})}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{x^2 -(x^2 + 8x)}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{-8x}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{\frac{-8x}{|x|}}{\frac{x}{|x|} -\sqrt{\frac{x^2}{|x|^2} + \frac{8x}{|x|^2}}}&& |\cdot |\text{ is needed since }x<0\\
&=\lim_{x\to -\infty}\frac{8}{-1 -\sqrt{1 - 8/x}}&&\text{since }|x|=-x\,\text{ for }x<0\\
&=\frac{8}{-1-\sqrt{1-0}}\\
&=\color{blue}{-4}
\end{align}
A: The problem is with your "simplify further". You are forgetting that $x<0$, and that $\sqrt{x^2}=|x|$. You have, since $|x|/x=-1$, 
$$
\frac{-8x}{x - \sqrt{{x^2} + 8x}}
=\frac{-8}{1 - \frac{\sqrt{{x^2} + 8x}}x}
=\frac{-8}{1 - \frac{|x|\sqrt{{1} + \frac8x}}x}
=\frac{-8}{1 + \sqrt{{1} + \frac8x}}\longrightarrow\frac{-8}2=-4.
$$
A: The limit as the question stands does not exist.
I assume you want to calculate the limit
$\lim_{x\to -\infty} x -\sqrt{x^2 + 8x}$
take $x=-y$
Now, you ca follow you procedure
$\lim_{y\to \infty} -y -\sqrt{y^2 + 8x}$
$=\lim_{y\to \infty}\frac{-8y}{\sqrt{y^2-8y}+y}$
$=\lim_{y\to \infty}\frac{-8}{\sqrt{y-8}+2}$
Now apply the limit, you get $-4$
A: Set $-1/x=y, x\to-\infty\implies y\to0^+, y>0$
$\sqrt{x^2+8x}=\sqrt{\dfrac{1-8y}{y^2}}=\dfrac{\sqrt{1-8y}}{|y|}=\dfrac{\sqrt{1-8y}}y$ as $y>0$
$$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}=\lim_{y\to0^+}\dfrac{-1+\sqrt{1-8y}}y =\lim_{y\to0^+}\dfrac{1-8y-1}{y(\sqrt{1-8y}+1)}=\cdots=\dfrac{-8}{\sqrt1+1}=?$$
