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What is the limit of $f(x)$ as $x \to 0$ if it exists? Justify your answers. $$f(x)= \begin{cases} \sin(x) & \text{if }x\notin\mathbb{Q} \\0 & \text{if }x\in \mathbb{Q} \end{cases}$$

What I did: Define two sequences $a_n=\frac{1}{n}$ and $b_n= 2\pi n -\frac{3\pi}{2}$.

Then $f(a_n) \to 0$ as $n \to \infty $ and $f(b_n) \to 1$ as $n \to \infty $. Hence this limit does not exist.

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  • $\begingroup$ how did you get $f(b_n) \to 1$ as $n \to \infty$? $\endgroup$ – pointguard0 Aug 12 '18 at 16:22
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    $\begingroup$ $b_n\not\to0 $. $\endgroup$ – Lord Shark the Unknown Aug 12 '18 at 16:22
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    $\begingroup$ Hint: The limit exists and is zero. $\endgroup$ – amsmath Aug 12 '18 at 16:23
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    $\begingroup$ I think $f(x)\rightarrow 0$ as $x\rightarrow 0$. $\endgroup$ – Lev Ban Aug 12 '18 at 16:25
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The answer is that you are not choosing a correct pair of sequences to prove that the limit doesn't exist, since the limit actually exists! It is equal to $0$. This follows from the squeeze theorem and the fact that$$(\forall x\in\mathbb{R}):\bigl\lvert f(x)\bigr\rvert\leqslant|x|.$$

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  • $\begingroup$ I understand now that I cannot use these sequences. I see how $f(x)→0$ when $x$ is rational. But when $x$ is irrational, I can deduce only $-1\leqslant f(x)\leqslant1$. How do you go from here? $\endgroup$ – user499701 Aug 12 '18 at 16:46
  • $\begingroup$ Did you read my answer? I provided a proof of the fact that $\lim_{x\to0}f(x)=0$. $\endgroup$ – José Carlos Santos Aug 12 '18 at 16:49
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    $\begingroup$ You have the inequality $-1\le \sin x\le 1$ which doesn't help to solve the problem. In the answer you can see one inequality ($|\sin x|\le |x|$) that solves the problem. $\endgroup$ – mfl Aug 12 '18 at 17:03

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