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?
 A: There exists, by the Mean Value Theorem, $c_x \in (0,x)$ such that $$\frac{1 - \cos x}{x}=\frac{1 -\cos x }{0 - x } = -\frac{d}{dx}\big(\cos c_x\big) = \sin c_x$$
thus $$\frac{1 - \cos x}{x} = \sin c_x$$
Consider $n\Big(1 - \cos \Big(\frac{1}{n}\Big)\Big) = \sin c_n$, where $ 0 < c_n < \frac{1}{n} $. Now there exists $d_x \in (0,x) $ such that 
$$\frac{0 - \sin x}{x}  = \frac{0 - \sin x}{0 - x}= -\cos d_x \geq -1  \Rightarrow \sin x \leq x$$
Then 
$$0 \leq n\Big(\underbrace {1 -\cos \Big(\frac{1}{n}}_{\geq 0}\Big)\Big) = \sin c_n \leq c_n < \frac{1}{n}$$
Applying the Squeeze Theorem we have 
$$\lim_{n \to \infty} n\Big(1 -\cos \Big(\frac{1}{n}\Big)\Big) = 0 $$
A: 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,
$$\lim_{n \to \infty} n (1-\cos(1/n))=0.$$
A: 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 $\cos t \le 1$.
A: 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.
A: Hint: How does the term $\frac{1 - \cos x}{x}$ relate to (A) your problem, (B) the mean value theorem?
A: Hint:
$$
n(1-\cos(1/n))=\frac1n\cdot\left(\frac{\sin(1/n)}{1/n}\right)^2\cdot\frac1{1+\cos(1/n)}
$$
