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!
 A: 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 $\mathbb R\cup\{\pm\infty\}$".
A: 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.
A: 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 $\liminf a_n = \limsup a_n = 0$.
