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!

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}
