When I calculate the limit $$\lim_{x\to\infty}\sqrt{x^2+x} - \sqrt {x^2-x}$$ I get $2$ answers for this question: $1$ and $0$ but $1$ is the right answer. I don't know why this is the case, however. If you multiply by the conjugate divided by the conjugate (1), you take the radical out of top and get it in the bottom and then if you factor out $x$ from both and cancel it with top u get $2/2$ which is $1$. But if you just factor you get $$\lim_{x\to\infty} x (\sqrt {1+ 1/x} - \sqrt{1 - 1/x}.$$ This simplifies to 0. So how would you know which method to use if you didn't know the right answer?
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2$\begingroup$ "But if you just factor you get $lim_{x \to +\infty} x (\sqrt{1+ 1/x} - \sqrt{1 - 1/x})$. Which simplifies to 0." - It doesn't. $\endgroup$– AbstractionDec 13, 2015 at 20:18
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$\begingroup$ @NormalHuman hahahahhaa I missed it. $\endgroup$– Jimmy R.Dec 13, 2015 at 20:22
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4 Answers
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You are right when multiplying the conjugate
$$\sqrt{x^2+x} - \sqrt {x^2-x}={2x\over\sqrt{x^2+x} + \sqrt {x^2-x}}={2\over\sqrt{1+{1\over x}} + \sqrt {1-{1\over x}}}$$
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1$\begingroup$ Be careful that the limit has different sigh for $x\to +\infty$ or $x \to -\infty$. The correct result is $\dfrac{2}{sign(x)\left(\sqrt{1+{1\over x}} + \sqrt {1-{1\over x}} \right)}$ $\endgroup$ Dec 13, 2015 at 20:31
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$ x\bigl( \sqrt{1+\tfrac1x} - \sqrt{1-\tfrac1x} \bigr) $ as $x\to\infty$ is of the form $\infty\cdot0$, which is indeterminate hence your wrong conclusion. A correct way to get rid of the indeterminate form is as you suggested earlier: indeed, \begin{align} \sqrt{x^2+x} - \sqrt{x^2-x} ={}& \frac{ \left(\sqrt{x^2+x} - \sqrt{x^2-x}\right) \left(\color{red}{\sqrt{x^2+x} + \sqrt{x^2-x}}\right) }{ \color{red}{\sqrt{x^2+x} + \sqrt{x^2-x}} } \\ ={}& \frac{2x}{ \sqrt{x^2+x} + \sqrt{x^2-x} } \\ ={}& \frac{2x}{ x\left(\sqrt{1+\frac1x} + \sqrt{1-\frac1x}\right) } \\ ={}& \frac{2}{ \sqrt{1+\frac1x} + \sqrt{1-\frac1x} } \xrightarrow{x\to+\infty} \frac 22 = 1 \end{align} By doing so you passed from an indeterminate to a well definite form.
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for any constant $0 \lt C \in \mathbb{R}$ we have $$ \left|\lim_{r\to\infty} (\sqrt{r^2-C}- r)\right| \lt \left| \lim_{r\to\infty}\frac{-C}{r} \right|=0 $$ hence $$ \lim_{x\to\infty} ( \sqrt{x^2+x}-\sqrt{x^2-x}) -1\\ = \lim_{x\to\infty} \left( \left( \sqrt{(x+\frac12)^2-\frac14}- (x+\frac12) \right) -\left( \sqrt{(x-\frac12)^2-\frac14}- (x-\frac12) \right)\right) =0 $$ so $$ \lim_{x\to\infty} ( \sqrt{x^2+x}-\sqrt{x^2-x}) =1 $$
answered Dec 13, 2015 at 20:54
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But if you just factor you get $\lim_{x\to\infty} x (\sqrt {1+ 1/x} - \sqrt{1 - 1/x}).$
This simplifies to 0.
The difference of square roots is approximately equal to $1/x$, so that the expression inside the limit is close in value to $x (1/x) = 1$. It converges to that value as $x \to \infty$.
Where you wrote "simplifies to 0" that must mean "simplifies, in the limit, to 0" or "converges to 0" or something like that. The difference of square roots is not equal to $0$ for any finite $x$.
So how would you know which method to use if you didn't know the right answer?
I can't say how one would be guaranteed to avoid trouble and get correct answers using computational basic calculus methods only. But noticing that the form of the limit is $\infty \cdot 0$ might raise the suspicion that some care has to be taken, and maybe it can then be solved by some standard subroutine like Hopital's rule or rationalizing the fractions.
