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$$ \lim_{x\to 0+} \sin\left(\frac{1}{x}\right)$$ I know that there is no limit.

but, why there is no limit? I tried $x=0.4$, $x=0.3$, $x=0.1$, it looks like the limit is $0$.

And how can I show that there is no limit? I tried to calculate it like all the other functions, and I got wrong result and I don't know why:

$$\lim_{x \to 0+} \sin\left(\frac{1}{x}\right) = \sin\left(\frac{1}{0^+}\right) = \sin\left(\frac{1}{\infty}\right) = \sin(0) = 0.$$

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    $\begingroup$ Since when is $0^+=\infty$? $\endgroup$ – user228113 Dec 30 '15 at 10:39
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    $\begingroup$ How come $\frac{1}{0+}=1/\infty$?Did you get where you went wrong? $\endgroup$ – Suraj_Singh Dec 30 '15 at 10:41
  • $\begingroup$ You cannot substitute $0^+$ and $\infty$ as if they were numbers, because they actually aren't. $\endgroup$ – rubik Dec 30 '15 at 10:45
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    $\begingroup$ Try taking the limit along the sequence $x_k = \frac{1}{k \pi/2}$ with $k$ integer. $\endgroup$ – Winther Dec 30 '15 at 10:45
  • $\begingroup$ I would be highly surprised if this question isn't a duplicate. $\endgroup$ – MathematicsStudent1122 Dec 30 '15 at 12:06
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Why there is no limit?

The graphic can help you understand why and suggest you some approach for the proof:

enter image description here

Remark: You have to be careful with tables of values because they can be misleading:

\begin{array}{ c | c c c c } x & \frac{1}{2\pi} & \frac{1}{3\pi} & \frac{1}{4\pi} &\frac{1}{5\pi} \\ \hline \sin\left(\frac{1}{x}\right) & 0 & 0 & 0 & 0 \\ \end{array}

\begin{array}{ c | c c c c } x & \frac{2}{5\pi} & \frac{2}{9\pi} & \frac{2}{13\pi} &\frac{2}{17\pi} \\ \hline \sin\left(\frac{1}{x}\right) & 1 & 1 & 1 & 1 \\ \end{array}

(The tables above are a sketch of the proof - see Theorem 2.4 here.)

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Hint: $\frac1x$ has a right-sided limit of $+\infty$. Does $\sin x$ have a limit for $x\to\infty$?

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You made an error. \begin{array}{ c | c c c c } x & .4 & .3 & .2 &.1 \\ \hline \sin\left(\frac{1}{x}\right) & 0.5984721 & -0.190568 & -0.9589243 & -0.54402113 \\ \end{array} This does not suggest that the limit approaches $0$.

Second, you mean to say $$\lim_{x\to 0^+} = \sin\left(\frac{1}{x}\right) = \sin(\infty)$$ not $\lim_{x\to 0^+}\sin\left(\frac{1}{x}\right) = \sin\left(\frac{1}{\infty}\right).$ Recall that $$\lim_{x\to 0^+} \frac{1}{x} = \infty.$$

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Consider the two subsequences $x_k = \frac1{(4k + 1)\pi/2}$ and $x'_k = \frac1{k\pi}$ with $k \in \mathbb Z$. We have that for $k \to +\infty$, $x_k,x'_k \to 0^+$. Clearly, $$\begin{align*} &\lim_{k \to +\infty} \sin\left(\frac1{x_k}\right) = 1\\ &\lim_{k \to +\infty} \sin\left(\frac1{x'_k}\right) = 0 \end{align*}$$ and therefore the limit $x \to 0^+$ does not exist. We used the theorem that states that if a sequence converges, then every subsequence converges to the same limit. By modus tollens, our sequence does not converge.

More info about the theorem here: Prove: If a sequence converges, then every subsequence converges to the same limit.

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That's because your limit is equivalent to $$\lim_{x\to\infty} \sin(x)$$

and $\sin$ is oscillating when $x\to\infty$.

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