Let $f(x)= \cos(x/2)\cdot\cos(x/4)\cdots\cos(x/2^n)$.If lim $ \lim_{n \to infinity} f(x) =g(x)$ , such that $\lim_{x \to 0} g(x)=k$, then the value of $\lim_{k \to 1} (1-k^{2011})/(1-k)$
I want to know that my solution is correct or not. If it incorrect please tell me where I went wrong.
There is an error in the OP.
Letting $\theta=x/2^{k}$ in the double-angle formula $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ reveals
$$\cos(x/2^{k})=\frac{\sin(x/2^{k-1})}{2\sin(x/{2^k})} \tag 1$$
for $x\ne 0$.
Then, using $(1)$, we can write
$$\begin{align} f_n(x)&=\prod_{k=1}^n \cos(x/2^{k})\\\\ &=\prod_{k=1}^n \frac{\sin(x/2^{k-1})}{2\sin(x/{2^k})}\\\\ &=\left(\frac{\sin(x)}{2^n \sin(x/2^{n})}\right)\\\\ \end{align}$$
for $x\ne 0$ with $f_n(0)=1$. Then, taking the limit we find that
$$\lim_{n\to \infty} f_n(x)=\frac{\sin(x)}{x}$$
for all $x$. If we designate the limit by $g(x)$, then clearly $g(x)=\frac{\sin(x)}{x}$ for all $x\ne 0$ and $g(0)=0$.
Therefore, $\lim_{x\to 0}g(x)=1$.
On a seemingly unrelated limit, we see that
$$\frac{1-k^{2011}}{1-k}=\sum_{n=0}^{2010} k^n$$
so that
$$\begin{align} \lim_{k\to 1} \left(\frac{1-k^{2011}}{1-k}\right)&=\lim_{k\to 1}\sum_{n=0}^{2010} k^n\\\\ &=\sum_{n=0}^{2010} (1)\\\\ &=2011 \end{align}$$
