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Find the following limit: $$ \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )$$

I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$

When I plug in the $\infty$, I'm left with $ \frac{\infty}{\infty} $. Did I mess up somewhere, or does the limit not exist?

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    $\begingroup$ Since you have $\infty/\infty,$ looks like a job for $l'Hopital.$ Have you tried it? $\endgroup$ – Igor Rivin Dec 7 '13 at 19:13
  • $\begingroup$ I'm aware of the rule, but this is part of review for earlier units in my Calculus class. I don't think my professor wants us to use that on this particular question. $\endgroup$ – Ahounsel Dec 7 '13 at 19:21
  • $\begingroup$ Realted : math.stackexchange.com/questions/596558/… $\endgroup$ – lab bhattacharjee Dec 8 '13 at 4:10
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You are almost there:

$$\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{x \sqrt{1+\frac{1}{x}}+x\sqrt{1-\frac{1}{x}}}$$

so...

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Hint:

$$ax^2+bx+c\approx ax^2, ~~x\to\infty$$ so if $x\to+\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=x$ and if $x\to-\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=-x$

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For $x\ge 1$ we have $$ \sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}. $$ It follows that $$ \lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1. $$

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Change it with $t=1/x$ (for $x>1$), which transforms it into $$ \lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt{1-t}}{t} $$ Now choose whatever method you're familiar with:

  • multiplying by $\sqrt{1+t}+\sqrt{1-t}$

  • recognize the derivative at $0$ of $f(t)=\sqrt{1+t}-\sqrt{1-t}$

  • Taylor expansion at degree $1$

If the limit is at $-\infty$, it's better to use $t=-1/x$, in order to avoid sign mistakes.

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I'm going to show a heuristic that can give you an idea of what you want to show. It's not rigorous, but often, it's easier to find the rigorous demonstration once you know what the likely result is.

The general idea is that $(x+a)^2 = x^2+2ax+a^2$, and for constant $a$, we can ignore the $a^2$ term as $x \to \infty$. Thus, we can approximate $\sqrt{x^2+2ax}$ by $x+a$, and moreover, this approximation gets better as $x$ grows without bound.

In the case of this problem, we quickly obtain $\sqrt{x^2+x} \doteq x+\frac{1}{2}$, and $\sqrt{x^2-x} \doteq x-\frac{1}{2}$, so their difference is approximately $1$. That turns out to be the result. One can extend this notion rigorously or, as in the other answers, there is often an alternate approach that shows convergence more readily. But again, finding that approach is often made easier by knowing what the limit is likely to be.

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