Prove that the following sequence $(sinfrac{pi k}{3})_{k=1}^infty$ diverges

$$x_k := sin left(frac{pi k}{3} right),qquad x_{infty} := lim_{k to infty}{x_k}$$

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I tried to prove by contradiction. Assume limit exists and is $L$. Fix $epsilon=frac{sqrt{3}}{4}$. There exists $Kinmathbb{Z}^{+}$ such that $k>K$ implies $$|sinfrac{pi k}{3}-L|<epsilon$$ Notice that for $k=6l+1$ or $k=6l+2$ where $linmathbb{Z}^{geq 0}$, $lvert frac{sqrt{3}}{2}-Lrvert<frac{sqrt{3}}{4}$. Also, for $k=6l+4$ or $k=6l+5$ where $linmathbb{Z}^{geq 0}$, $lvert -frac{sqrt{3}}{2}-Lrvert<frac{sqrt{3}}{4}$. Now we know that $sqrt{3}=lvertfrac{sqrt{3}}{2}-big(-frac{sqrt{3}}{2}big)rvert=|frac{sqrt{3}}{2}+L-L-big(-frac{sqrt{3}}{2}big)|leq|frac{sqrt{3}}{2}+L|+|-L-big(-frac{sqrt{3}}{2}big)|$ $=|-frac{sqrt{3}}{2}-L|+|frac{sqrt{3}}{2}-L|<frac{sqrt{3}}{4}+frac{sqrt{3}}{4}=frac{sqrt{3}}{2};Rightarrow;sqrt{3}<frac{sqrt{3}}{2}$. Hence, we get a contradiction. Therefore, the sequence diverges.

Is this proof correct? If not, can you help me to prove using epsilon-delta definition?