Let we have the limit as :

$L=\lim _{ n\rightarrow \infty }{ \sqrt { n } { x }_{ n } }$

Also $L=\lim _{ n\rightarrow \infty }{ \sqrt { n+1 } { x }_{ n+1 } }$

Now removing the limits I am taking some n arbitarily large

Since n is way too large we have :

$\sqrt { n+1 } \approx \sqrt { n } (1+\frac { 1 }{ 2n })$

Also ${x}_{n+1}=sin({x}_{n})$ also since ${x}_{n}$ is way too small we have

$sin({x}_{n})\approx{x}_{n}-\frac { { x }_{ n }^{ 3 } }{ 6 }$

So using these we have

$L=\sqrt { n } { x }_{ n }(1+\frac { 1 }{ 2n } )(1-\frac { { x }_{ n }^{ 2 } }{ 6 } )$

Also $L=\sqrt { n } { x }_{ n }$ using this we get :

$1=1+\frac { 1 }{ 2n } -\frac { { x }_{ n }^{ 2 } }{ 6 }$

(Note : I have neglected the term $\frac { -{ x }_{ n }^{ 2 } }{ 12n }$ here because it is too small)

So we get :

$n{ x }_{ n }^{ 2 }=3={ L }^{ 2 }$

$\boxed { \Rightarrow L=\sqrt { 3 } }$

This problem is nice example of Stolz-Cesàro Theorem. For anyone, who don't know about this , here it is :

Let $(X_n)$ , $(Y_n)$ be two sequences that satisfy the conditions:

• $(Y_n)$ strictly increases to $+\infty$ ,
• $\lim\limits_{n\to\infty} \frac{X_n-X_{n-1}}{Y_n-Y_{n-1}}=a$

Then $\lim\limits_{n\to\infty} \frac{X_n}{Y_n}=a.$

Now , note that $\lim\limits_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}$ Therefore $\lim\limits_{n\to\infty}\left(\frac1{x_{n+1}^2}-\frac1{x_n^2}\right)=\frac{1}{3}$

Now, taking $Y_n=n$ and, $X_n = \frac{1}{x_n^2}$ and using Stolz-Cesàro Theorem, we have $\lim\limits_{n\to\infty} \frac{1}{nx_n^2} =\frac{1}{3} \implies \lim\limits_{n\to\infty} \sqrt{n}x_n = \sqrt{3}$

I've used c++ programming to find the result by putting arbitrarily large value in input which gave me a good approximation about 1.728.

here's my program..

"

include<iostream.h>

include<math.h>

int main() { float x=1; float a[10000]; float c=1; int n=9000; for(int i=1; i<=n;i++) { a[0]=1; a[i+1]=sin(a[i]); x=a[i+1]; } c=sqrt(n)*x; cout<<c; return 0; } try it out//.