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?

Obviously, $$\left|1-\cos \left(\frac{1}{n} \right) \right| = \left|\cos(0)-\cos \left(\frac{1}{n} \right) \right|.$$ By the mean value theorem there exist $a_n \in (0,1/n)$ such that
$$\left|1-\cos \left(\frac{1}{n} \right) \right| = |\cos'(a_n)| \left( \frac{1}{n} - 0 \right) = \sin(a_n) \frac{1}{n}.$$
Since $\sin(0)=0$ and $x \mapsto \sin(x)$ is continuous, we have
$$a_n \to 0 \quad \text{as} \, n \to \infty \implies \sin(a_n) \to 0 \quad \text{as} \, n \to \infty.$$
Combining both facts, we get
$$\begin{align*} 0 &\leq \liminf_{n \to \infty} \left| n (1-\cos(1/n)) \right| \\ &\leq \limsup_{n \to \infty} \left| n (1-\cos(1/n)) \right| \\ &\leq \limsup_{n \to \infty} \sin(a_n) = 0. \end{align*}$$
(In the last step, we have used that $\limsup = \lim$ whenever the limit exists.) Consequently,
$$\lim_{n \to \infty} n (1-\cos(1/n))=0.$$
