How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?
 A: Exploit the identity:
$$\sqrt{x^2+6x}-\sqrt{x^2-2x} = \frac{8x}{\sqrt{x^2+6x}+\sqrt{x^2-2x}} = \frac{8x}{2|x|+O(1)}.$$
A: There's a subtle point when you algebraically manipulate the expression (this can be seen in several of the given answers): since you can assume that $x<0$ (actually $x<-6$ in order that the expression is meaningful), you have to remember that
$$
\sqrt{x^2}=|x|=-x
$$
So
\begin{align}
\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})
&=\lim_{x\to-∞}\frac{(x^2+6x)-(x^2-2x)}{\sqrt{x^2+6x}+\sqrt{x^2-2x}}\\
&=\lim_{x\to-∞}\frac{8x}{\sqrt{x^2(1+6/x)}+\sqrt{x^2(1-2/x)}}\\
&=\lim_{x\to-∞}\frac{8x}{|x|\bigl(\sqrt{(1+6/x)}+\sqrt{(1-2/x)}\,\bigr)}\\
&=\lim_{x\to-∞}\frac{8}{-\bigl(\sqrt{(1+6/x)}+\sqrt{(1-2/x)}\,\bigr)}\\
&=-4
\end{align}
A: Set $-\dfrac1x=h$ to
$$x^2+6x=\frac{1-6h}{h^2}\implies\sqrt{x^2+6x}=\frac{\sqrt{1-6h}}{\sqrt{h^2}}$$
and as $x\to-\infty,h\to0^+$ and $h>0,\sqrt{h^2}=+h$
So we have
 $$\lim_{x\to-\infty}(\sqrt{x^2+6x}-\sqrt{x^2-2x})=\lim_{h\to0^+}\frac{\sqrt{1-6h}-\sqrt{1+2h}}h$$
$$=\lim_{h\to0^+}\frac{1-6h-(1+2h)}{h(\sqrt{1-6h}+\sqrt{1+2h})}$$
A: $$\sqrt{x^2+6x}=\sqrt{x^2+6x+9-9}=\sqrt{(x+3)^2-9}\approx |x|+3,\text{ as }x\to-\infty$$
$$\sqrt{x^2-2x}=\sqrt{x^2-2x+1-1}=\sqrt{(x-1)^2-1}\approx |x|-1,\text{ as }x\to-\infty$$
A: Note that $$ \sqrt{x^2+6x}-\sqrt{x^2-2x} = \frac{8x}{\sqrt{x^2+6x}+\sqrt{x^2-2x}}, $$so our answer is the limit of the above as $ x \to -\infty $. But note that $ \sqrt {x^2 + 6x} + \sqrt {x^2 - 2x} \sim \mathcal{O}(2x) $, so the answer is $\frac{8}{2}=\boxed{4}$. 
A: To conclude this The answer is actually -4 , because you can do this step 

