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I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below. $$ \lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right) $$

I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick. That doesn't seem to work. I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :(

Thanks.

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  • $\begingroup$ Probably one of the most duplicated questions, under one guise or another. Hint: conjugate quantity. $\endgroup$ – Did Feb 13 '14 at 20:36
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    $\begingroup$ Note that $\sqrt{x^2} = \big| x \big|$. In particular, when $x < 0$ (which you consider when $x \to -\infty$), the correct equation is $\sqrt{x^2} = -x$. $\endgroup$ – Sammy Black Feb 13 '14 at 20:38
  • $\begingroup$ yes. thanks. i see it now. $\endgroup$ – Bill Feb 13 '14 at 20:45
  • $\begingroup$ Intuitively, this equals $\lim_{x\to-\infty}(\sqrt{x^2+2x+1}-\sqrt{x^2-2x+1})$, which is trivial except for the signs. $\endgroup$ – Yves Daoust May 13 '16 at 15:27
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The direct approach of just factoring an $|x|$ from each piece is not fruitful: It leads to

$$ |x| \Big(\sqrt{1 + 2/x} - \sqrt{1 - 2/x}\Big)$$

The first term grows, and the second term tends to $0$, so there's a balance between them.


Multiply top and bottom by the conjugate to find that

\begin{align*} \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} &= \Big(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}\Big) \left(\frac{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}\right) \\ &= \frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}} \\ &= \frac{-4}{\sqrt{1 + 2/x} + \sqrt{1 - 2/x}} \end{align*}

since $\sqrt{x^2} = |x| = -x$ for $x < 0$. Can you finish it from here?

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  • $\begingroup$ ok. I can do it now. i guess i had brain lock. thank you. $\endgroup$ – Bill Feb 13 '14 at 20:42
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    $\begingroup$ $x\to -\infty$, so you get $$\frac{-4}{\sqrt{1+2/x} + \sqrt{1-2/x}}.$$ $\endgroup$ – Daniel Fischer Feb 13 '14 at 20:45
  • $\begingroup$ @DanielFischer Yes, I was missing the sign; thanks. $\endgroup$ – user61527 Feb 13 '14 at 20:49
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For $x>0$: For brevity let $A=\sqrt {x^2+2 x}.$ We have $(x+1)^2=A^2 +1>A^2>0$ so $x+1>A>0 . $ ..... So we have $$0<x+1-A=$$ $$=(x+1-A)\frac {x+1+A}{x+1+A}=\frac {(x+1)^2-A^2}{x+1+A}=\frac {1}{x+1+A}<1/x.$$ Therefore $$(i)\quad \lim_{x\to \infty} (x+1-A)=0.$$ For $x>2$: For brevity let $B=\sqrt {x^2-2 x}.$ We have $(x-1)^2=B^2+1>B^2>0$ so $x-1>B>0 . $ ..... So we have $$0<x-1-B=$$ $$=(x-1-B)\frac {x-1+B}{x-1+B}=\frac {(x-1)^2-B^2}{x-1+B}=\frac {1}{x-1+B}<1/(x-1).$$ Therefore $$(ii)\quad\lim_{x\to \infty}(x-1-B)=0.$$

We have $\sqrt {x^2+2 x}-\sqrt {x^2-2 x}=$ $A-B=(A-(x+1))-(B-(x-1))+2. $ From $(i)$ and $(ii)$ we have $$\lim_{x\to \infty} A-B=\lim_{x\to \infty}(A-(x+1))-(B-(x-1))+2=0+0+2=2.$$

The idea is that when $x$ is large, $A$ is close to $x+1$ and $B$ is close to $x-1$ so $A-B$ is close to $(x+1)-(x-1)=2.$

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So:

$$ \lim_{x\to-\infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}) = \lim_{x\to-\infty}(\frac{x^2 + 2x - x^2 + 2x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{-4}{\sqrt{1 + \frac{2}{x}} + \sqrt{1 - \frac{2}{x}}}) = -2 $$

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