# 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
1 year, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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.

- 1 year, 3 months ago

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

- 1 year, 3 months ago

- 1 year, 3 months ago

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/

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

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

Thank You!

- 1 year, 3 months ago

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

- 1 year, 3 months ago

This has nothing to do with abstract algebra.

- 1 year, 3 months ago