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Prove that $f\left( x\right) =\begin{cases} \sin \dfrac {1} {x},x \neq 0\\ 0,x=0\end{cases}$

has no limit as $x\rightarrow 0$.

Proof. Consider $a_{n}=\dfrac {2} {\left( 4n+1\right) \pi }$ and $b_{n}=\dfrac {2} {\left( 4n+3\right) \pi }$, $n\in\mathbb{N}$.

Clearly, both $a_{n}$ and $b_{n}$ converge to $0$ as $n\rightarrow \infty$. On the other hand, since $f\left( a_{n}\right) =1$ and $f\left( b_{n}\right) =-1$ for all $n\in\mathbb{N}$, $f\left( a_{n}\right) \rightarrow 1$ and $f\left( b_{n}\right) \rightarrow -1$ as $n\rightarrow \infty$. Thus, the limit of $f(x)$, as $x\rightarrow \infty$, cannot exist.

My question is: Why consider $a_{n}=\dfrac {2} {\left( 4n+1\right) \pi }$ and $b_{n}=\dfrac {2} {\left( 4n+3\right) \pi }$?

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    Because it happens to work? – 2017-01-23
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    Because $\;\lim\limits_{x\to0}f(x)\;$ exists iff exists $\;\lim\limits_{n\to\infty}f(x_n)\;$ , for **any sequence** $\;x_n\xrightarrow[n\to\infty]{}0\;$ . They just chose two different sequences converging to zero and for which the limit of $\;f(x_n)\;$ is different... – 2017-01-23
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    As opposed to another choice of sequences or what is the underlying technique used? If you ask why this approach is used, @DonAntonio provides an answer. If you ask why these particular sequences are used, well Arthur does. So we have a rare case of 2 comments providing a full answer. :-) – 2017-01-23
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    @Arthur Okey. So, How did writer think this? – 2017-01-23
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    @Kahler He was familiar with the input values that makes the regular sine function into $\pm1$, and he tailored $a_n$ and $b_n$ such that $1/a_n$ hits its maximum points and $1/b_n$ its minimum points. – 2017-01-23
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    @Arthur So, can we consider more different $a_n$ and $b_n$? If yes, can you write here? – 2017-01-23
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    We could use $c_n = \frac{1}{\pi n}$, for instance, and compare to either $a_n$ or $b_n$. Or we could use $d_n = 1/n$ by itself and show that the resulting sequence $f(d_n)$ doesn't have a limit at all. There are loads of options. – 2017-01-23
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    @Arthur Could we use $a_{n}=\dfrac {2} {n\pi }$ and $b_{n}=\dfrac {2} {3n\pi }$? – 2017-01-23
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    @Kahler In that case, only $a_n$ is enough, because $f(a_n)$ goes $1, 0, -1, 0, 1,0\ldots$, which doesn't have a limit. But yes, you could use both of them if you wanted. – 2017-01-23
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    @Arthur I understood, thanks. – 2017-01-23

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