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Please how can I show that $$\displaystyle\lim_{x \rightarrow 0} \frac{1}{x} \sin\frac{1}{x}$$ does not exist?

I tried the substitution $u = \frac{1}{x}$ so that I'm now calculating $\lim_{u \rightarrow + \infty} u \sin(u)$ But from there I don't know what to do.

Thanks in advance

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  • $\begingroup$ what does the graph of $u\sin u$ look like? $\endgroup$
    – operatorerror
    Jan 2, 2018 at 14:20

2 Answers 2

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Take $$x_n=\frac{1}{2\pi n}\quad \text{and}\quad y_n=\frac{1}{\frac{\pi}{2}+2\pi n},$$

and compute $$\lim_{n\to \infty }f(x_n)\quad \text{and}\quad \lim_{n\to \infty }f(y_n).$$

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  • $\begingroup$ I get zero in one case and Π/2 in the other so the limit doesn't exist thanks $\endgroup$
    – arsene stein
    Jan 2, 2018 at 14:32
  • $\begingroup$ By the way how are we supposed to guess which sequences to use? $\endgroup$
    – arsene stein
    Jan 2, 2018 at 14:36
  • $\begingroup$ You want to fix the sinus, the way to do it is to take such sequences. $\endgroup$
    – idm
    Jan 2, 2018 at 14:50
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You're on the right track. Suppose $\lim_{u\to\infty}u\sin u = L$, where $L$ is some number (i.e., the limit exists). Note that the function $u\sin u$ oscillates between positive and negative values, and for $u > 1$ we can find values with magnitude $> 1$. So we can find infinite values $ < -1$ as $u \to \infty$, and we can find infinite values $>1$ as $u\to\infty$. This will not allow any value of $L$.

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  • $\begingroup$ OK thanks it's clear this way $\endgroup$
    – arsene stein
    Jan 2, 2018 at 14:36

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