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Prove that the limit of sin(x) as it approaches infinity does not exist using the formal definition of a limit.

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Let $\epsilon=1$ (or, if you wish, something smaller, like $1/4$). If there was a number $a$ such that $\lim_{x\to\infty} \sin x=a$, then there would be a $B$ such that $|\sin x-a|\lt \epsilon$ for all $x\gt B$.

But there are arbitrarily large $s$ such that $\sin s=1$, and also arbitrarily large $t$ such that $\sin t=-1$. Show that $a$ cannot be simultaneously at distance $\lt 1$ from $1$ and from $-1$. Formally, this part of the argument uses the Triangle Inequality.

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Suppose it does exist and seek a contradiction. It shouldn't be too difficult to find something inconsistent with your proposed limit and the oscillatory nature of sine.

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Suppose that you use the limit of sequence to define the limit of function.

To prove that $\displaystyle \lim_{x\rightarrow\infty}\sin(x)$ does not exist we choose two different real sequences $\{u_n\}$ and $\{v_n\}$ such that $ \lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}v_n=\infty, $ and $ \lim_{n\rightarrow\infty}\sin(u_n)\ne\lim_{n\rightarrow\infty}\sin(v_n). $ For example we choose $ u_n=\frac{\pi}{2}+n2\pi, \quad v_n=\pi+n2\pi \quad (n\in \mathbb{N}). $ Then $ \lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}v_n=\infty, $ and $ \lim_{n\rightarrow\infty}\sin(u_n)=1\ne0=\lim_{n\rightarrow\infty}\sin(v_n). $