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

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

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

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

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

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

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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). – Brian Charlesworth · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

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

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. – Brian Charlesworth · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

Thanks! I understood nowLog in to reply

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

\[ \lim_{x \rightarrow 0 } \frac{1}{x} = \text{ undefined }, \lim_{ x \rightarrow 0 } x \times \frac{1}{x} = 1 \text{ is defined } \] – Calvin Lin Staff · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

Thanks! I got it now.Log in to reply

@Brian Charlesworth , @Calvin Lin , @Prasun Biswas , @Sandeep Bhardwaj ...help please... – Akhil Bansal · 1 year, 9 months ago

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