Limit of $\cos(n\,t)/n$ does exist and converge under conditions? I thought $\displaystyle{\lim_{n\to 0} \dfrac{\cos(n)}{n}}$ does not exist, but under some conditions I noticed, it does:
$c_n = \dfrac{n^2\,\pi ^2\,\sin\left(\pi \,n\right)-2\,\sin\left(\pi \,n\right)+2\,\pi \,n\,\cos\left(\pi \,n\right)}{n^3\,\pi }$
Here, that term converges to $\frac{\pi^2}{3}$
It arose from $\displaystyle{c_n = \dfrac{1}{2\,\pi}\int_{-\pi}^{\pi}t^2\,e^{-i\,n\,t}}$ (Fourier integral)
It is somewhat clear that integral also is defined for $n = 0$. I just can't figure out why.
$\displaystyle{\left(\text{As stated in the headline:} \quad\lim_{n\to 0}\dfrac{2\,\cos\left(\pi \,n\right)}{n^2\,\pi} \quad \text{shouldn't exist on its own.}\right)}$
 A: Let's rewrite this as
$$c_n = \frac{n^2\pi^2\sin\left(\pi n\right)}{n^3\pi} - \frac{2\sin\left(\pi n\right)}{n^3\pi}+\frac{2\pi n\cos\left(\pi n\right)}{n^3\pi}.$$
The first term of that sum $\frac{n^2\pi^2\sin\left(\pi n\right)}{n^3\pi} = \frac{\pi^2\sin\left(\pi n\right)}{\pi n}$ goes to $\pi^2$ as $n \to 0$.
The second term $-\frac{2\sin\left(\pi n\right)}{n^3\pi} = -\frac2{n^2}\frac{\sin\pi n}{\pi n}$ goes to $-\infty$ as $n\to 0$, and it does so asymptotically like $-\frac2{n^2}$.
The third term, $\frac{2\pi n\cos\left(\pi n\right)}{n^3\pi} = \frac2{n^2}\cos(\pi n)$, goes to $+\infty$ as $n \to 0$, and it does so asymptotically like $\frac2{n^2}$.
So if you split the sum into 3 parts, one part will converge, but 2 parts diverge. While the first summand is non-problematic, the other 2 are and they cannot be handled in separation, that just gives you an indeterminate form of $-\infty + \infty$, and you need to evaluate that limit in some other way than "simply" adding the individual limits, as they don't exist (as you noticed for the third part).
In a nutshell, you discovered a complicated version of $\lim_{n\to 0}\left( \frac1{\frac1{n}} - \frac1{\frac1{n+1}}\right) = -1$ and wondered how that can be when $\lim_{n\to 0}\frac1{\frac1{n}}$ obviously doesn't exist. That's correct, but if you subtract another term that grows roughly the same way, you might get a term that actually has a limit, as the 2 growing terms cancel each other and the remaining difference is small and converges.
