Does $\lim_{n \to \infty} n \cos(n^2 x)$ exist for some $x$? (Michael Spivak "Calculus 3rd Edition" p.497) I am reading Michael Spivak "Calculus 3rd Edition".  
On p.497 Spivak wrote as follows:  

if $$f_n(x) = \frac{1}{n} \sin(n^2 x),$$ then $\{f_n\}$ converges uniformly to the function $f(x) = 0$, but $$f_n^{'}(x) = n \cos(n^2 x),$$ and $\lim_{n \to \infty} n \cos(n^2 x)$ does not always exist(for example, it does not exist if $x = 0$).

Then, 
does $\lim_{n \to \infty} n \cos(n^2 x)$  exist for some $x$?
 A: If $\lim_{n\to\infty} n\cos(n^2x)$ exists then $\cos(n^2x)\to 0$ and hence $$n^2x=(2a_n+1)\frac{\pi}{2}+b_n$$ where $a_n\in\mathbb {Z}, b_n\to 0$. Replacing $n$ by $n-1$ and subtracting the resulting equation from the last equation we get $$(2n-1)x=c_n\pi+d_n$$ where $c_n\in\mathbb {Z}, d_n\to 0$. Applying the same argument again we see that $$2x=e_n\pi+f_n$$ where $e_n\in\mathbb {Z}, f_n\to 0$. Taking limits we see that $e_n\to 2x/\pi$. Since $e_n$ takes integer values it follows that the sequence $e_n$ ultimately becomes a constant, say $k$, and thus $x=k\pi/2$ for some integer $k$.
The case $k=0$ when $x=0$ is simplest one and has been mentioned by Spivak. For other values of $k$ the sequence $\cos(n^2x)$ oscillates or tends to $1$ (check this). And hence $n\cos(n^2x)$ never tends to a limit for any given value of $x$. 
A: First, we observe that if $\cos n^2x \neq 0$, then $f_n$ does not have a limit, since $n \cos n^2x$ increases in magnitude.
Therefore, we look for cases where $\cos n^2 x$ can be zero so that the resulting sequence is the constant zero sequence, which will have a limit.
First, notice that if $n$ is odd, then $\cos n^2 x$ is zero for $x = \dfrac{\left( 2k + 1 \right) \pi}{2}$, where $k \in \mathbb{Z}$. However, for the same $x$, if $n$ is even, $\cos n^2x \neq 0$. Therefore, at this value of $x$, the sequence will not have a limit because it will have a seubsequence which increases in magnitude.
Similarly, if we try to make $\cos n^2x$ zero at even $n$, then the odd $n$'s create similar problems. Therefore, the sequence under consideration can never have a limit.
