# Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$

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.

• thanks Semiclassical-typo on my part – Shreyas B. Jun 18 '16 at 22:17
• First, multiply by $\sqrt{x^2+1}+x\over \sqrt{x^2+1}+x$. – David Mitra Jun 18 '16 at 22:22
• @ShreyasB. Use $(a-b)(a+b) = a^2 - b^2$ and multiply by $(1/x)\over(1/x)$ – A---B Jun 18 '16 at 22:27
• Beware of old Russian books, otherwise you end up with problems like these....+1 – imranfat Jun 18 '16 at 23:09

In this case, consider what would happen if you multiply both numerator and denominator by $\sqrt{x^2+1} + x$.
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.
• I might be missing something but did you drop a square root sign? I.e shouldn't it be $1/[\sqrt (1 + x ^{-2})+1]? Which is the same result of course. – fleablood Jun 18 '16 at 22:50 • @fleablood Thanks. I fixed the error. – Foobaz John Jun 18 '16 at 23:05 Hint: Let$t=1/x^2$and recall the definition of the derivative as a difference quotient. One interesting thing to note is that:$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}$is the asymptote quite easily.$\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}$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