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?

If $x > 0$, apply the mean value theorem to the cosine to get $1 - \cos x = \sin(c_x)x$ where $0 < c_x < x$. Then $1 - \cos(1/n) = \sin(c_n)/n$ for all $n\in \Bbb N$, where $c_n$ is a sequence such that $0 < c_n < 1/n$. Thus, $$n(1 - \cos(1/n)) = \sin(c_n) \le c_n \le \frac{1}{n} \quad (n\in \Bbb N).$$ From this, deduce that $\lim_{n\to \infty} n(1 - \cos(1/n)) = 0$.
Note: To establish $\sin(c_n) \le c_n$ for all $n$, use the mean value theorem for $\sin x$, $x > 0$, and the inequality $\cos t \le 1$.
