If $\frac{\sin{a_n}}{a_n}\rightarrow 1$ then $a_n\rightarrow0$

I'm trying to prove that if $\;\frac{\sin{a_n}}{a_n}\rightarrow 1$ then $a_n\rightarrow0$, assuming $a_n\neq 0$ for all $n$.

I think this is easy enough to show as follows: first, prove $f(x)=\frac{\sin(x)}{x},f(0)=1$ has a global maximum at 0, then assume by negation that not $a_n\rightarrow0$, and reach a contradiction with epsilon-delta gymnastics. But this will turn out to be a long and somewhat messy proof.

Is there a more elegant way to prove this?

Addendum: this isn't a homework question (I'm not sure if it looks like one), so if possible, please give the full details of your answer.

Thanks!

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Let $f:\mathbb R\cup\{\pm\infty\}\to\mathbb R$ be the unique continuous function which coincides with $(\sin x)/x$ on $\mathbb R\setminus\{0\}$.

Then $f$ takes the value $1$ at $0$, and nowhere else.

This implies that any subsequence of $(a_n)$ which converges in $\mathbb R\cup\{\pm\infty\}$ converges to $0$, and thus, that the sequence $(a_n)$ itself converges to $0$.

EDIT A. More details on the last point: Assume by contradiction $(a_n)$ does not converge to $0$. Then there is a positive $\varepsilon$ and a subsequence $(b_n)$ such that $|b_n|\ge\varepsilon$ for all $n$, and $(b_n)$ admits in turn a subsequence $(c_n)$ which converges in $\mathbb R\cup\{\pm\infty\}$. The limit of $(c_n)$ cannot be $0$, contradiction.

EDIT B. The reason for working not in $\mathbb R$ but in $\mathbb R\cup\{\pm\infty\}$ is that any sequence in $\mathbb R\cup\{\pm\infty\}$ has a converging subsequence. A topological space is said to be sequentially compact if it has this property. The fact that $\mathbb R\cup\{\pm\infty\}$ is sequentially compact is used above in the phrase "$(b_n)$ admits in turn a subsequence $(c_n)$ which converges in $\mathbb R\cup\{\pm\infty\}$".

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Thanks! This looks good. –  ro44 Jan 10 '12 at 6:54
Dear @ro44: Thank you! +1 for your nice question! –  Pierre-Yves Gaillard Jan 10 '12 at 6:57
Pierre-Yves: 1. It seems the hypothesis of your proof by contradiction should be that $(a_n)$ does not converge to $0$. 2. One could expect the word compact to appear somewhere in the proof, maybe. 3. Nice answer. –  Did Jan 10 '12 at 7:51
Dear @Didier: Thanks! I edited the answer. The previous formulation was due to the fact that the edit was a reply to a (now deleted) comment from the OP asking something like "How do you know that $(a_n)$ converges?". It was clear that the OP knew that the limit could only be $0$. In fact I was about to make this change. But without your comment, I wouldn't have thought of mentioning compactness explicitly - which was a good idea. –  Pierre-Yves Gaillard Jan 10 '12 at 8:35

Look at the function $f(x) = \frac{\sin(x)}{x}$, extended continuously to $x=0$ with $f(0) = 1$ (see here). Clearly, if you pick a small $\epsilon > 0$, then the only way $|f(x) - 1| < \epsilon$ is if $x$ is sufficiently close to $0$, i.e. if $-\delta(\epsilon) < x < \delta(\epsilon)$ where $\delta(\epsilon)$ is a small number depending on $\epsilon$. Notice that $\delta(\epsilon) \rightarrow 0$ as $\epsilon \rightarrow 0$. Since $f(a_n) \rightarrow 1$, for every $\epsilon > 0$ there will be an $N$ so that if $n \geq N$ then $|f(a_n) - 1| < \epsilon$, which means $-\delta(\epsilon) < a_n < \delta(\epsilon)$ for all $n \geq N$. This shows that $\{a_n\}$ indeed converges to 0.

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Isn't the point of this question to prove what's hidden behind "Clearly, ..."? –  t.b. Jan 10 '12 at 5:51
@t.b.There's nothing "hidden". The function |f(x)| achieves its maximum on the interval $[-\pi,\pi]$ at $x = 0$, where it is equal to 1, and only there. Outside of that interval, we have $|f(x)| < 0.8$ (I'm too lazy to get a better upper bound that 0.8). That's it. –  user18063 Jan 10 '12 at 6:04
@user18063: Thank you for your answer. This is the proof I had in mind when phrasing the question. If you should fully write out the proof (without the assistance of the graph), it turns out to be somewhat long and messy, which is why I am dissatisfied with it. –  ro44 Jan 10 '12 at 6:19

Here is a similar approach, in which you do not need to show that the limit exists.

Let $A=\limsup a_n$, and suppose that $a_{n_k} \to A$.

Let $f(x) = \frac{\sin(x)}{x}$ with $f(0) = 1$.

As $f$ is continuous, we have that $1 = \lim_{k \to \infty} f(a_{n_k}) = f(A)$, so that $A=0$. Doing the same with $\liminf a_n$ we get that $\liminf a_n = \limsup a_n = 0$.

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