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Evaluate:

$$\lim_{x\to 0} \sin (\dfrac {1}{x})$$

My Attempt:

let $$\dfrac {1}{x}=k$$ As $\dfrac {1}{x}\to 0$, $k\to \infty$ Now, $$=\lim_{k\to \infty} \sin k$$

Nosrati
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pi-π
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    Hint: Plug in the sequence $x_n=2/(n\pi)$ for $x$ and check what happens when $n\to\infty$. – sranthrop Sep 21 '17 at 01:18
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    You are correct so far. Now all you have to notice is that the function $\sin(k)$ oscillates between $-1$ and $1$. Regardless of how far you go out, it will keep oscillating and will never approach one value. Therefore, the limit doesn't exist. – John Lou Sep 21 '17 at 01:20

2 Answers2

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HINT: you can consider the sequences $x_n = \frac{2}{n\pi}$, or $x_n=\frac{2}{4(n+1)\pi}$, or $x_n=\frac{2}{4(n-1)\pi}$ when $n\to \infty$.

GAVD
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You are correct in your reasoning thus far. You can observe that the limit of $\sin x$ as $x$ approaches infinity doesn't exist, and neither does this limit.

paulinho
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