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Problem: Show that $a_n$ is convergent sequence and find a limit of $a_n$. $$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$$ I tried to look at this as normal limit problem so I wrote this: $$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\lim_{n \to \infty}(\frac{1}{\sqrt{1+1}})=\frac{1}{2}$$ But I didn't get anything which can help me to solve a problem.

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    $\begingroup$ How is $a_n$ defined? Also $\frac{1}{\sqrt{2}} \neq \frac{1}{2}$ $\endgroup$ – Piotr Benedysiuk Feb 7 '16 at 16:10
  • $\begingroup$ Is $a_n >0$ perhaps? $\endgroup$ – fosho Feb 7 '16 at 16:14
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You can invert the function $y = {x \over \sqrt{x^2 + 1}}$ as follows. $$y^2 = {x^2 \over x^2 + 1} = 1 - {1 \over x^2 + 1}$$ $$1 - y^2 = {1 \over x^2 + 1}$$ $${1 \over 1 - y^2} = x^2 + 1$$ $${1 \over 1 - y^2} - 1 = x^2$$ So we have $$x^2 = {y^2 \over 1 - y^2}$$ Seeing that $x$ and $y$ must have the same sign, we have $$x = {y \over \sqrt{1 - y^2}}$$ Hence if for your sequence $x_n$ you write $y_n = {x_n \over \sqrt{1 + x^2}}$, then you have $$x_n = {y_n \over \sqrt{1 - y_n^2}}$$ Since $\lim_{n \rightarrow \infty} y_n = {1 \over 2}$, by the continuity of ${y \over \sqrt{1 - y^2}}$ you have $$\lim_{n \rightarrow \infty} x_n = {{1 \over 2} \over \sqrt{1 - {1 \over 4}}}$$ $$= {1 \over \sqrt{3}}$$

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  • $\begingroup$ You're faster than I. This is the solution approach I was going to post! +1 $\endgroup$ – Mark Viola Feb 7 '16 at 16:29
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    $\begingroup$ gotta have fast fingers in this game... :) $\endgroup$ – Zarrax Feb 7 '16 at 16:33

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