Application of the mean value theorem to find $\lim_{n\to\infty} n(1 - \cos(1/n))$ While reading Heuser (2009) "Lehrbuch der Analysis Teil I" on page 286, I got this question:
Find $$\lim\limits_{n \rightarrow \infty} n\Big(1 - \cos\Big(\frac{1}{n}\Big)\Big)$$ with the help of the Mean Value Theorem.
How do you apply the Mean Value Theorem to this problem?

Hint:
$$\frac{1-\cos x}{x} = \frac{2\sin^2(x/2)}{x} = \frac{x}{2}\cdot\left(\frac{\sin(x/2)}{x/2}\right)^2.\tag{1}$$
Since $\lim_{z\to 0}\frac{\sin z}{z}=1$ (by the MVT or whatsoever), what is the limit of the LHS of $(1)$ when $x\to 0$?
As an alternative, for any $x\in\mathbb{R}^+$,
$$0\leq\frac{1-\cos x}{x}=\frac{1}{x}\int_{0}^{x}\sin y\,dy\leq\frac{1}{x}\int_{0}^{x}y\,dy = \frac{x}{2},$$
so by letting $x\to 0^+$ we get that the limit is zero just as before.
