Find the limit of $x +\sqrt{x^2 + 8x}$ as $x\to-\infty$

Solution 1:

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}

Solution 2:

Notice, $$\lim_{x\to -\infty}(x+\sqrt{x^2+8x})$$ substituting $x=-x$ $$=\lim_{x\to \infty}(-x+\sqrt{x^2-8x})$$ $$=\lim_{x\to \infty}(\sqrt{x^2-8x}-x)\frac{(\sqrt{x^2-8x}+x)}{(\sqrt{x^2-8x}+x)}$$ $$=\lim_{x\to \infty}\frac{x^2-8x-x^2}{\sqrt{x^2-8x}+x}$$ $$=-8\lim_{x\to \infty}\frac{x}{x\sqrt{1-\frac{8}{x}}+x}$$ $$=-8\lim_{x\to \infty}\frac{x}{x\left(\sqrt{1-\frac{8}{x}}+1\right)}$$ $$=-8\lim_{x\to \infty}\frac{1}{\sqrt{1-\frac{8}{x}}+1}$$ $$=-8\left(\frac{1}{\sqrt{1-0}+1}\right)$$ $$=-\frac{8}{2}=\color{red}{-4}$$