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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?

Martin Sleziak
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taro
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    is this not the derivative of $\cos t $ at $t = 0?$ i guess you want to use MVT to derive this? it will $\sin(\theta /n)$ for $0 < \theta < 1$ and let $n \to \infty$ – abel Jan 10 '15 at 21:28
  • I think that @abel's comment deserves to be the accepted answer. It is so spot on and (I guess) very much in the spirit of Heuser... – Dirk Jan 11 '15 at 15:38
  • @Dirk, k.stm gave me the first hint and saz told me that $\cos(0) = 1$. I selected Aaron Maroja's answer as the selected answer because it looks most complete, though it could have been more concise. I feel abel's comment is surely concise and good, but sometimes it is so hard to read between lines when people try to explain as short as possible. – taro Jan 12 '15 at 15:07
  • Isn't the proof suggested by that textbook in a sense circular. When we use Mean-Value Theorem, we need to know that $(\cos x)'=\sin x$. But by definition $\lim\limits_{x\to 0} \frac{\cos x-1}x = (\cos x)'|_{x=0} = \sin 0 =1$, so (a more general version of) the limit we are trying to find is used in the proof that derivative of $\cos x$ is $\sin x$. – Martin Sleziak Jan 21 '15 at 06:10

6 Answers6

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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

$$\lim_{n \to \infty} n\Big(1 -\cos \Big(\frac{1}{n}\Big)\Big) = 0 $$

Aaron Maroja
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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,

$$\lim_{n \to \infty} n (1-\cos(1/n))=0.$$

saz
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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 $\cos t \le 1$.

kobe
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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.

Jack D'Aurizio
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    (+1) Both this and mine are based on $\lim\limits_{x\to0}\frac{\sin(x)}x$. I like that better than the mean value theorem since it has nice geometrical proofs. However, the question does ask for the mean value theorem. – robjohn Jan 11 '15 at 12:06
  • @robjohn: I just added a few lines proving the claim with the MVT (integral version). – Jack D'Aurizio Jan 11 '15 at 13:46
  • Indeed, $\lim\limits_{x\to0}\frac{\sin(x)}x$ can be evaluated using the MVT, and this is what is shown in kobe's answer (which I've also upvoted). – robjohn Jan 11 '15 at 14:45
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Hint: How does the term $\frac{1 - \cos x}{x}$ relate to (A) your problem, (B) the mean value theorem?

k.stm
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Hint: $$ n(1-\cos(1/n))=\frac1n\cdot\left(\frac{\sin(1/n)}{1/n}\right)^2\cdot\frac1{1+\cos(1/n)} $$

robjohn
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