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I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$ Could someone please explain to my why this is the case?

Thanks a lot.

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  • $\begingroup$ thanks Semiclassical-typo on my part $\endgroup$
    – Shreyas B.
    Commented Jun 18, 2016 at 22:17
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    $\begingroup$ First, multiply by $\sqrt{x^2+1}+x\over \sqrt{x^2+1}+x$. $\endgroup$ Commented Jun 18, 2016 at 22:22
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    $\begingroup$ @DavidMitra Please put answers as answers, not as comments. $\endgroup$
    – user231101
    Commented Jun 18, 2016 at 22:24
  • $\begingroup$ @ShreyasB. Use $(a-b)(a+b) = a^2 - b^2$ and multiply by $(1/x)\over(1/x) $ $\endgroup$
    – user312097
    Commented Jun 18, 2016 at 22:27
  • $\begingroup$ Beware of old Russian books, otherwise you end up with problems like these....+1 $\endgroup$
    – imranfat
    Commented Jun 18, 2016 at 23:09

6 Answers 6

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When you get questions like this (i.e. a one-line limit with a radical), always consider multiplying both numerator and denominator by "something" to make it look nicer.

In this case, consider what would happen if you multiply both numerator and denominator by $\sqrt{x^2+1} + x$.

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  • $\begingroup$ Got it, thanks GrancescoFrechi and David Mitra! $\endgroup$
    – Shreyas B.
    Commented Jun 18, 2016 at 22:26
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Mutiplying the numerator and denominator by $\sqrt{x^2+1}+x$ we get that $$\begin{align}\lim_{x\to \infty} x(\sqrt{x^2+1} - x) &=\lim_{x\to \infty} \frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}\\ &=\lim_{x\to \infty} \frac{x}{|x|\sqrt{1+x^{-2}}+x}\\ &=\lim_{x\to \infty} \frac{1}{\sqrt{1+x^{-2}}+1} \end{align}$$ where we have used the fact that $|x|=x$ because the limit is as $x$ goes to infinity. From here the limit is easy to compute.

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  • $\begingroup$ I might be missing something but did you drop a square root sign? I.e shouldn't it be $ 1/[\sqrt (1 + x ^{-2})+1]? Which is the same result of course. $\endgroup$
    – fleablood
    Commented Jun 18, 2016 at 22:50
  • $\begingroup$ @fleablood Thanks. I fixed the error. $\endgroup$ Commented Jun 18, 2016 at 23:05
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Hint: Let $t=1/x^2$ and recall the definition of the derivative as a difference quotient.

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One interesting thing to note is that:

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

Implies that

$x=\frac{k}{\sqrt{1-2k}}$

And you can then see how $k=\frac{1}{2}$ is the asymptote quite easily.

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

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

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

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Another approach based on Taylor expansions.

Consider $$y=x(\sqrt{x^2+1} - x)=x^2\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ and remember that, for small $y$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$. Replace $y$ by $\frac{1}{x^2}$ to get $$\sqrt{1+\frac{1}{x^2}}-1=\frac{1}{2 x^2}-\frac{1}{8 x^4}+O\left(\frac{1}{x^5}\right)$$ which makes $$y=\frac{1}{2}-\frac{1}{8 x^2}+O\left(\frac{1}{x^3}\right)$$ showing the limit and also how it is approched

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