How can I prove that $$\lim_{x\to \infty}{\sin(2x)}$$ does not exist?
How can I prove this with the epsilon-delta definition?
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1 Answer
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We can prove by contradiction that the limit does not exist. Assume the limit $a$ exists: $$ \forall \epsilon>0 \exists\delta > 0 : x>\delta \Rightarrow |\sin 2x - a| < \epsilon$$
But this would also mean that $$ \forall \epsilon>0 \exists\delta > 0 : x_1,x_2>\delta \Rightarrow |\sin 2x_1-\sin 2x_2| < 2\epsilon$$
Use $\epsilon=1$ and pick $x_1 = N\pi+\frac{\pi}{4}, x_2 = N\pi+\frac{3\pi}{4}$ where $N$ is large enough so that $x_1,x_2 > \delta$. Now we have $\sin 2x_1 = 1$ and $\sin 2x_2 = -1$, which means $|\sin 2x_1 - \sin 2x_2| = 2 > \epsilon = 1$
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