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$$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$ plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?

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6 Answers 6

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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}

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Exploit the identity: $$\sqrt{x^2+6x}-\sqrt{x^2-2x} = \frac{8x}{\sqrt{x^2+6x}+\sqrt{x^2-2x}} = \frac{8x}{2|x|+O(1)}.$$

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  • $\begingroup$ Beat me to it! +1 $\endgroup$ Dec 25, 2014 at 9:41
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    $\begingroup$ I am stuck on the third step how did u turn the square roots to 2x + 0(1)? $\endgroup$
    – method
    Dec 25, 2014 at 9:47
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    $\begingroup$ $$\sqrt{x^2+Ax} = x\sqrt{1+\frac{A}{x}} = x\cdot\left(1+O\left(\frac{1}{x}\right)\right) = x+O(1).$$ $\endgroup$ Dec 25, 2014 at 9:54
  • $\begingroup$ To be more precise, $$x+\frac{A}{2}\geq\sqrt{x^2+Ax}\geq x+\frac{A}{2}-\frac{A^2}{8x}$$ by a similar technique. $\endgroup$ Dec 25, 2014 at 9:58
  • $\begingroup$ I don't fully understand what you did, on the calculator square root of squaredx + ax is not equal to x, I see that you took x^2 common then got it out as x what did u do next? $\endgroup$
    – method
    Dec 25, 2014 at 10:00
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$$\sqrt{x^2+6x}=\sqrt{x^2+6x+9-9}=\sqrt{(x+3)^2-9}\approx |x|+3,\text{ as }x\to-\infty$$ $$\sqrt{x^2-2x}=\sqrt{x^2-2x+1-1}=\sqrt{(x-1)^2-1}\approx |x|-1,\text{ as }x\to-\infty$$

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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 $$\lim_{x\to-\infty}(\sqrt{x^2+6x}-\sqrt{x^2-2x})=\lim_{h\to0^+}\frac{\sqrt{1-6h}-\sqrt{1+2h}}h$$

$$=\lim_{h\to0^+}\frac{1-6h-(1+2h)}{h(\sqrt{1-6h}+\sqrt{1+2h})}$$

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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 $\frac{8}{2}=\boxed{4}$.

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  • $\begingroup$ Really? A negative function has positive limit? $\endgroup$
    – egreg
    Dec 25, 2014 at 10:44
  • $\begingroup$ The issue is that $\sqrt{x^2+6x}+\sqrt{x^2-2x}=2|x|+O(1)$. $\endgroup$ Dec 25, 2014 at 12:03
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To conclude this The answer is actually -4 , because you can do this step

and the answer is clearly 4.

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  • $\begingroup$ No, the limit is $-4$. $\endgroup$
    – egreg
    Dec 25, 2014 at 10:44

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