Question 238695
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\,\sin\left(\frac{1}{x}\right)]


As *[tex \Large x] approaches 0, *[tex \Large \frac{1}{x}] increases without bound. Thus in any open interval containing 0 there will be values of *[tex \Large x] such that *[tex \Large \frac{1}{x}] is a multiple of *[tex \Large 2\pi], values of *[tex \Large x] such that *[tex \Large \frac{1}{x}] is *[tex \Large \frac{\pi}{2}] more than a multiple of *[tex \Large 2\pi], and values of *[tex \Large x] such that *[tex \Large \frac{1}{x}] is *[tex \Large \frac{3\pi}{2}] more than a multiple of *[tex \Large 2\pi]. The sine of these values of *[tex \Large x] will be 0, 1 and -1 respectively.


*[tex \Large \sin\left(\frac{1}{x}\right)] oscillates "an infinite number of times" between 1 and -1 in any neighbourhood of *[tex \Large x\ \rightarrow\ 0]. Therefore the limit doesn't exist.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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