Finding $\lim\limits_{n→∞}n\cos x\cos(\cos x)\cdots\underbrace{\cos(\cos(\cdots(\cos x)))}_{n\text{ times of }\cos}$ 
Find$$\lim_{n→∞}n\cos x\cos(\cos x)\cdots\underbrace{\cos(\cos(\cdots(\cos x)))}_{n \text{ times of } \cos}.$$

I approximated cos(cosx) to cos x, but i don't think it is the proper approach.
 I got answer as 0 on the approximation.
It is clear that it is a 0/0 form, but how can the l's Hopital rule be applied? I tried using the sandwich theorem but I am unable to reach the answer.
I plotted the graph on desmos. But I got the resultant graph covering the entire area. 
please help me reach the proper answer. Thanks in advanced to all. 

Let $x_0$ denote the root of $\cos(x) = x$, then we have $x_0 \approx 0.739085$. Define
$$\cos_n(x) = \underbrace{\cos(\cos(\cdots(\cos x)))}_{n \text{ times of} \cos}.$$
First, we can prove that for all $x \in \mathbb{R}$,
$$
\begin{align}
\lim_{n \to \infty} \cos_n(x) = x_0.\tag{1}
\end{align}
$$
From (1), for any given $x$, there exists $N_0(x) > 0$, such that for all $k > N_0(x)$ we have
$|\cos_k(x)| < \frac{x_0 + 1}{2}$. Therefore, for all $n > N_0(x)$, 
$$
\begin{align}
0 \leq \left|n \prod_{k = 1}^{n} \cos_k(x) \right| &< n \left|\prod_{k = 1}^{N_0(x)} \cos_k(x)\right| \left(\frac{x_0 + 1}{2}\right)^{n - N_0(x)}\\ &= \left|\prod_{k = 1}^{N_0(x)} \cos_k(x)\right| \left(\frac{x_0 + 1}{2}\right)^{-N_0(x)} \cdot n \,\left(\frac{x_0 + 1}{2}\right)^n.
\end{align}
$$
Since $ \left|\frac{x_0+1}{2}\right| < 1$, from the squeeze theorem we have
$$
\lim_{n \to \infty} n \prod_{k = 1}^{n} \cos_k(x) = 0.
$$

To prove (1) note that the range of cosine function is $[-1, 1]$, and from Lagrange's mean value theorem we have
$$
\begin{align}
\left|\cos_n(x) - x_0\right| &= \left|\cos(\cos_{n-1}(x)) - \cos(x_0)\right| \\
&\leq \sin(1) \cdot |\cos_{n-1}(x) - x_0| \leq \dots \leq \sin^{n-1}(1) \cdot \left|\cos(x) - x_0\right|.
\end{align}
$$
