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# Not defined * 0 = defined

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

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

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

Note by Akhil Bansal
1 year, 5 months ago

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While 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). · 1 year, 5 months ago

But if we combine both the steps,then they are not following the rules of maths. · 1 year, 5 months ago

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. · 1 year, 5 months ago

Thanks! I understood now · 1 year, 5 months ago

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

$\lim_{x \rightarrow 0 } \frac{1}{x} = \text{ undefined }, \lim_{ x \rightarrow 0 } x \times \frac{1}{x} = 1 \text{ is defined }$ Staff · 1 year, 5 months ago

Thanks! I got it now. · 1 year, 5 months ago