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Can this limit be cancelled?

So I have this limit here: \[ \lim_{x\to 0 } \dfrac{\sin \frac1x} { \sin \frac1x } \]

I know that the limit of just sin(\(\frac{1}{x}\)) (as x tends to 0) does not exist, as it is constantly oscillating between 1 and -1. My question is whether or not the fraction in the limit shown can be cancelled or not, as at some point the fraction can equal \(\frac{0}{0}\), which is not defined. What I really want to know is WHY it can or cannot be cancelled. If it is also possible, I would like to find out the answer to the limit too.

Thanks.

Note by Muhammed Li
5 months, 2 weeks ago

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Limit does not exist.

This is equivalent to evaluating the limit \( \displaystyle \lim_{y\to \infty} \dfrac{\sin y}{\sin y} \).

If \(y = \pi n\), for any positive integer \(n\), then \(\dfrac{\sin y}{\sin y} \) is always undefined, so there exists a subsequence such that this expression does not converge.

In other words: If it converges, then all its subsequences also converges. But since there is at least one subsequence that doesn't converge, then the limit does not converge.

Relevant article: Subsequences.

Pi Han Goh - 5 months, 2 weeks ago

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Thank you for explaining it to me. I was looking for a well expounded and concise explanation and I got one :)

Muhammed Li - 5 months, 1 week ago

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I think the limit of this function will come out to be 1, hence we can cancel both the term. At any value of x, the numerator and denominator will have the same value whose range will be between -1 and 1 .

I am expecting the community to give more insights on this topic. https://brilliant.org/profile/calvin-8u8hog/

Terry Chadwick - 5 months, 2 weeks ago

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I agree with Pi Han Goh's answer. If a subsequence does not converge, (in this case isn't even well-defined), the original sequence does not converge, i.e., the given limit does not exist.

Brian Charlesworth - 5 months, 2 weeks ago

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But, if y isn't a multiple of pi, then can't 1 be the answer as both numerator and denominator will cancel out?

I also found Pi Han Goh's answer great.

Terry Chadwick - 5 months, 2 weeks ago

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@Terry Chadwick For the limit to exist, every subsequence must converge, but the subsequence \(a_{n} = n \pi\) is not well-defined and thus does not converge. So while \(\frac{\sin(y)}{\sin(y)} = 1\) for all but a countably infinite number of points, the points \(n \pi\) where it does not equal \(1\) form an infinite sequence that does not converge and thus dictates the non-convergence of the given limit.

Brian Charlesworth - 5 months, 2 weeks ago

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@Brian Charlesworth Yeah, you are right. Though that this limit will exist for all real x except the ones whose multiple is pi.

Thank You!

Terry Chadwick - 5 months, 2 weeks ago

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I stand by those whose answer doesn't exist. I think according to abstract algebra algebra cancellation law holds only for non zero and known finite (also unique) numbers

Amit Kumar - 5 months, 1 week ago

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This has nothing to do with abstract algebra.

Pi Han Goh - 5 months, 1 week ago

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