For an arbitrary $x_{0}$ in $\left(\, 0,\pi\,\right)$ we define $x_{n + 1}=\sin\left(\, x_{n}\,\right)$.

Using the limit of the sequence as $n$ tends to infinity we're supposed to find the limit of $\,\sqrt{\,n\,}\,\sin\left(\,x_{n}\,\right)$ .

I literally don't know where to start, any ideas ?.

  • $\begingroup$ Hint is can Use Stolz $\endgroup$
    – math110
    Jan 7, 2015 at 5:32

1 Answer 1


Easily prove that $\lim x_n=0$. Then use L'Hôpital Rule, we have $$\frac{1}{x_{n+1}^2}-\frac{1}{x_n^2}=\frac{x_n^2-\sin^2 x_n}{x_n^2\sin^2 x_n}\rightarrow \frac{1}{3}$$ Then by Césaro theorem: $$\lim \frac{1}{nx_n^2}=\frac{1}{3}$$ so $\lim \sqrt{n}x_n=\sqrt{3}$


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