\[\large \displaystyle\lim_{x\rightarrow 0^{+}} \sin\left(\dfrac{1}{x}\right) = \ \text{Not defined}\]

\[\large \displaystyle\lim_{x\rightarrow 0} x\sin\left(\dfrac{1}{x}\right) = 0\]

Then how could \([\text{not defined} \times 0 = 0]\) ?

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## Comments

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TopNewestWhile the first limit is not defined, it is nevertheless bounded, as it oscillates between \(-1\) and \(1.\) Thus

\(-x \le x\sin\left(\dfrac{1}{x}\right) \le x,\)

and so by the Sandwich rule the second limit goes to \(0\) as \(x \rightarrow 0,\) (from both the left and the right).

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But if we combine both the steps,then they are not following the rules of maths.

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We're not really "combining" the steps; we are looking at two different problems. The first limit is undefined since the bounded oscillation of \(\sin(\frac{1}{x})\) never ceases or diminishes as \(x \rightarrow 0.\) However, in the second limit this oscillation, since it is bounded and multiplied by \(x\), diminishes to \(0\) as \(x \rightarrow 0,\) i.e., the diminution of \(x\) "beats out" the (bounded) oscillation of \(\sin(\frac{1}{x}),\) and so the product, and hence the limit, goes to \(0.\)

Some instances of "undefined times \(0\)" will be undefined, but in this case, because of the bounded nature of the "undefined" element, we do in fact find that the limit is defined.

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Shouldn't be surprising. Same holds when you remove the \(\sin\) function, IE

\[ \lim_{x \rightarrow 0 } \frac{1}{x} = \text{ undefined }, \lim_{ x \rightarrow 0 } x \times \frac{1}{x} = 1 \text{ is defined } \]

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Thanks! I got it now.

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@Brian Charlesworth , @Calvin Lin , @Prasun Biswas , @Sandeep Bhardwaj ...help please...

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