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Prove that the following limit does not exist.

$$ \lim_{x\to 0} \sin\left(1 \over x\right) $$

Our definition of a limit:

Let $L$ be a number and let ${\rm f}\left(x\right)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for ever $\epsilon > 0$ there exists a corresponding $\delta > 0$ such that $0 < \left\vert\,x-c\,\right\vert \left\vert\,{\rm f}\left(x\right) - L\,\right\vert$

Not really sure how I go about doing this?

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Doubleposted: math.stackexchange.com/questions/534364/ –  Rasmus Oct 21 '13 at 20:12

3 Answers 3

up vote 3 down vote accepted

Hint: Consider $x$ of the form

$$x_n = \frac{1}{2n\pi + \frac{\pi}{2}}$$

with $n$ a positive integer, and then consider

$$y_n = \frac{1}{2n\pi}$$

Then consider, say, $\epsilon = \frac{1}{2}$. Then for any $\delta$, there is an $n$ large enough that $x_n$ and $y_n$ are $\delta$-close to $0$.

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You Can use sequential Criterion to show limit does not exists by showing two different sequence converging to $0$ but the functional sequences does not converges to the same limit

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Suppose $k=\frac{1}{x}$. Your question would be the same as:

$\lim_{k\to\infty} \sin{(k)}$

Sine oscillates between -1 and 1, and does not have a limit

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