= Solution
Use $\sin0=0$, the <derivative> identity $\sin'=\cos$, <continuity> of cosine at zero, and $\cos0=1$. For $u\ne0$, the <mean value theorem> gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to1$ as $u\to0$. For $x\ne0$, apply this with $u=x/3^k$ to obtain \b[the limit] $\boxed{3^k\sin(x/3^k)\to x}$. For $x=0$ every term is zero.
The additional bound $|\cos t|\leq1$ and the same <mean value theorem> give $|\sin u|\leq|u|$. Therefore
$$
2^n\left|\sin\left(\frac x{3^n}\right)\right|\leq|x|\left(\frac23\right)^n.
$$
The right side is summable. By the <comparison test for series> \b[the stated series has <absolute convergence> for every real $x$].
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