Let \(x_0 , x_1, x_2,\ldots \) be a sequence of numbers satisfying the recursion \( x_{n+1} =\sqrt{\dfrac{1+x_n}2} \) for \(n=0,1,2,\ldots\) and \( -1 \leq x_0 \leq 1 \).

Find the value of \( \displaystyle \lim_{n\to\infty} \cos \left(\dfrac{\sqrt{1-x_0^2 }}{x_1 x_2 x_3\cdots x_n} \right) \).

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TopNewestLet \(x_0=\cos\theta_0\) for some \(\theta_0\in\Bbb R\) and note that \(-1\leq x_n\leq 1\implies 0\leq x_{n+1}\leq 1~\forall~n\geq 0\) using the recurrence relation which shows that \(0\leq x_n\leq 1~\forall~n\geq 1\).

Then, substitute \(x_n=\cos\theta_n~\forall~n\geq 1\) and using the recurrence relation, we get that \(x_{n+1}=|\cos(\theta_n/2)|~\forall~n\geq 0\) from which we conclude using the base case that \(x_n=|\cos(\theta_0/2^n)|~\forall~n\geq 1\). So, the limit in question becomes,

\[L=\lim_{n\to\infty}\cos\left(\frac{|\sin\theta_0|}{\prod\limits_{k=1}^n|\cos(\theta_0/2^k)|}\right)=\lim_{n\to\infty}\cos\left(\left|\frac{\sin\theta_0}{\prod\limits_{k=1}^n\cos(\theta_0/2^k)}\right|\right)\]

Multiplying both numerator and denominator by \(\sin(\theta_0/2^n)\) and using the sine double-angle formula repeatedly, the limit reduces to,

\[L=\lim_{n\to\infty}\cos\left(\left|\frac{\sin\theta_0}{\dfrac{\sin\theta_0}{2^n\sin(\theta_0/2^n)}}\right|\right)=\lim_{n\to\infty}\left(\left|\theta_0\cdot\frac{\sin(\theta_0/2^n)}{\theta_0/2^n}\right|\right)\]

Because of the continuity of the cosine and the modulus function, we can shift the limit operator inside the function body, hence, the limit becomes,

\[L=\cos\left(\left|\lim_{n\to\infty}\theta_0\cdot\frac{\sin(\theta_0/2^n)}{\theta_0/2^n}\right|\right)\]

Now, substitute \(u=\theta_0/2^n\), so since \(\theta_0\) is a finite real number, \(u\to 0\) as \(n\to\infty\). Then, the limit becomes,

\[L=\cos\left(\left|\theta_0\lim_{u\to 0}\frac{\sin u}{u}\right|\right)=\cos\left(|\theta_0|\right)=\cos\theta_0=x_0\]

In the second step in the above line, we used the well-known result that \(\lim\limits_{x\to 0}\frac{\sin x}{x}=1\) which can be easily proved using the squeeze theorem.

So, we conclude that,

\[\lim_{n\to\infty}\cos\left(\frac{\sqrt{1-x_0^2}}{x_1x_2\cdots x_n}\right)=x_0\] – Prasun Biswas · 8 months, 2 weeks ago

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please help me to solve this question. thanx in advance – Aaron Jerry Ninan · 8 months, 3 weeks ago

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