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. something like this please help me reach the proper answer. Thanks in advanced to all.

  • $\begingroup$ I edited the expression with MathJax. Please check if it's what you wrote. $\endgroup$ May 12, 2020 at 4:35
  • $\begingroup$ Yes, thanks for that! $\endgroup$
    – Aatmaj
    May 12, 2020 at 4:36
  • $\begingroup$ If $x=0$ then $\cos x=1$. $\cos 1\neq 0$ and I don't think that repeating $\cos$ will approach $0$, so I think the limit goes to $\infty$... $\endgroup$
    – abiessu
    May 12, 2020 at 4:48
  • $\begingroup$ But no constraint is given on x, and we cannot assume x=0 $\endgroup$
    – Aatmaj
    May 12, 2020 at 4:51
  • 2
    $\begingroup$ It is amazing that nobody mentioned Dottie number $\endgroup$ May 12, 2020 at 5:52

3 Answers 3


Consider the sequence $x_n$ defined by $x_0 = x$ and $x_{n+1} = \cos(x_n)$. Then the sequence in question is $$a_n = n\prod_{n=1}^\infty x_n.$$ I claim that $a_n \to 0$. Here's a sketch of the proof:

  1. There is a unique point $x^* \in [0, 1)$ such that $\cos(x^*) = x^*$ (the fixed point of $\cos$).
  2. The sequence of fixed point iterates $x_n$ converge to $x^*$, regardless of the value of $x$.
  3. The sum $\sum a_n$ converges, using the ratio test.
  4. The sequence $a_n$ converges to $0$, using the divergence test.

The hard bit is 2, which I'll leave to last. To prove 1, note that the function $f(x) = x - \cos(x)$ is continuous, negative at $0$, and positive at $\pi/2$, so by the intermediate value theorem, there must be at least one point where $f(x) = 0$ in $[0, \pi/2]$.

Further, $f'(x) = 1 + \sin(x) \ge 0$, meaning that the function is non-decreasing. If $f$ had more than one root, then it'd be an interval of roots, which would correspond to an interval of roots in $f'$. This is clearly not the case, so there is a unique $x^*$ such that $f(x^*) = 0$, i.e. $\cos(x^*) = x^*$.

The point $x^*$ lies in the range of $\cos$, i.e. $[-1, 1]$, as well as $[0, \pi/2]$, so $x^* \in [0, 1]$. If $x^* = 1$, then $\cos(x^*) = 1$, hence $x^*$ would have to be an integer multiple of $2\pi$, which it is clearly not. Thus, $x^* \in [0, 1)$, as claimed.

To prove 3, assuming 2 is proven, consider $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{n}|x_{n+1}| \to x^* < 1,$$ thus the series converges absolutely. Then, 4 follows immediately from this: the terms of a convergent series must tend to $0$.

Now, we tackle 2. First, recall the trigonometric identity: $$\cos(x) - \cos(y) = -2\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right).$$ Now, suppose that $x, y \in [0, 1]$. Note that $\frac{x + y}{2} \in [0, 1]$ and $\sin$ is increasing and positive on $[0, 1] \subseteq [0, \pi/2]$, hence $$\left|\sin\left(\frac{x + y}{2}\right)\right| = \sin\left(\frac{x + y}{2}\right) \le \sin(1).$$ Also, recall that $|\sin \theta| \le |\theta|$ for all $\theta$. Hence, assuming still $x, y \in [0, 1]$, $$|\cos(x) - \cos(y)| = 2\left|\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)\right| < 2 \cdot \sin(1) \cdot \left| \frac{x - y}{2}\right| = \sin(1)|x - y|.$$

Now, note that $x_n \in [-1, 1]$ for $n \ge 1$, and since $\cos$ is positive over $[-1, 1]$, we have $x_n \in [0, 1]$ for $n \ge 2$. So, for $n \ge 2$, we get $$|x_{n+2} - x_{n+1}| = |\cos(x_{n+1}) - \cos(x_n)| \le \sin(1)|x_{n+1} - x_n|.$$ This implies the series $$\sum_{n=2}^\infty (x_{n+1} - x_n)$$ is absolutely summable, as it passes the ratio test (limsup version): $$\left|\frac{x_{n+2} - x_{n+1}}{x_{n+1} - x_n}\right| = \frac{|x_{n+2} - x_{n+1}|}{|x_{n+1} - x_n|} \le \sin(1) \frac{|x_{n+1} - x_n|}{|x_{n+1} - x_n|} = \sin(1) < 1.$$ Therefore, $\sum_{n=2}^\infty (x_{n+1} - x_n)$ converges. This is a telescoping series, whose partial sums take the form $x_n - x_2$. These partial sums converge, and hence so must $x_n$.

Now, because $x_n$ converges to some $L$, it follows from $\cos$ being continuous that $$x_{n+1} = \cos(x_n) \implies L = \cos(L) \implies L = x^*,$$ completing step 2, and the full proof, as necessary.


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 $$ \begin{align} \left|\cos_n(x) - x_0\right| &= \left|\cos(\cos_{n-1}(x)) - \cos(x_0)\right| \\ &\leq \sin(1) \cdot |\cos_{n-1}(x) - x_0| \leq \dots \leq \sin^{n-1}(1) \cdot \left|\cos(x) - x_0\right|. \end{align} $$


We can say few more things. With the same notations as before: consider the sequence $x_{n+1} = \cos(x_n)$ with $0 < x_0 \leq 1$, and $x_*= \cos(x_*)$.

Heuristically, $x_n \to x^*$ very quickly. Thus, $x_0 x_1 ... x_n$ behaves like a geometric sequence : $$x_0 x_1 ... x_n \propto x_*^n \simeq 0.73^n \to 0$$

(consequently $n^k x_0 x_1 ... x_n \to 0$ for all $k > 0$)

To be more precise, consider: $$y_n := \frac1{x_*^{n+1}}x_0 x_1 ... x_n = \frac{x_0}{x_*} \frac{x_1}{x_*} ... \frac{x_n}{x_*}> 0$$

It's better to consider $\log y_n$ : $$\log y_n = \sum_{i = 0}^n \log(\frac{x_i}{x_*})$$ This série is really cool. First, the sign of $(-1)^i \log(\frac{x_i}{x_*})$ is constant. Indeed, if $x_n < x_*$ then $x_{n+1} > x_*$ : $$x_{n+1} - x_* = \cos(x_n) - \cos(x_*)$$

In fact, this série is geometric : $$|\log(\frac{x_i}{x_*})| = |\log(x_i) - \log(x_*)| = |\log(\cos(x_{i-1})) - \log(\cos(x_*))| \leq k |x_{i-1} - x_0|$$ with $k = \displaystyle\sup_{[0, 1]} |\tan| < \infty$. And, with a Taylor expansion : $$|\log(\frac{x_i}{x_*})| \leq k (\cos(1))^{i-1} |x_1 - x_0|$$

Thus, $y_n$ converges.


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