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- anonymous

I have the following function f(x)=sin(1/x) I've been told that the function is continuous for any x. I understand that the above function is continuous at (0,infinite) and also at (-infinite,0) but as much as I know the limit of this function while x -->0 does not exist or is undefined so how come the function is still continuous??

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- anonymous

- jamiebookeater

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- terenzreignz

Perhaps it's for all x in its domain?

- experimentX

the function is not continuous at x=0, are you missing something?

- anonymous

it's for \[f(x) \rightarrow R:R\]

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- experimentX

if f(x) = x sin(1/x), the function could have been made continuous ... you might be interested in http://math.stackexchange.com/questions/199603/is-exp-left-1-over-x2-right-differentiable-at-x-0

- anonymous

I might be missing some thing but I'm not sure what
I would have uploaded the question but it's in Hebrew

- anonymous

that I know. that's because x while x-->0 equals 0 and sin(1/x) is a Bounded function

- anonymous

sry I think that the right way to translate the question is: does the Intermediate value theorem is true for sin(1/x) and if it does (and it does) explain

- experimentX

bounded does not necessarily mean it is continuous. as x approaches to 0 the function oscillates violently between -1 and 1 without converging to any particular value. I don't think sin(1/x) is continuous at 0.
Both because limit and value are not exactly defined.

- anonymous

that's what I thought

- experimentX

f(x) = x sin(1/x) is particularly interesting ... being continuous but not differentiable at x=0.

- anonymous

but before I was talking about the following statement:
when given the following limit \[\lim_{x \rightarrow 0} f(x)*g(x) = ?\]
if \[\lim_{x \rightarrow 0} f(x) = 0\] AND g(x) is bounded then
\[\lim_{x \rightarrow 0} f(x)*g(x) = 0\]

- anonymous

which is the case that you're presenting

- experimentX

yeah ... the case is different from what you are asking.

- anonymous

but lets go back to this for a sec : does the Intermediate value theorem is true for sin(1/x) and if it does (and it does) explain
because if think this is the correct translation to the question I was given

- experimentX

the intermediate value theorem only applies for continuous function on some interval. Though for this particular Q i can say exactly that f(0) is between -1 and 1.

- anonymous

right that's the thing the title of the question was is the Intermediate value theorem correct only for continuous function?? here are a few functions etc. etc.

- anonymous

I think I understood the whole idea behind the question and I don't want to take too much of your time but any way thank you for your help :)

- experimentX

let me ask this to experts!! I'll reply back later.

- anonymous

tnk u

- experimentX

http://chat.stackexchange.com/transcript/message/6829811#6829811

- anonymous

yes I know there's no f(0) that's why the function is not continuous but why is it true that the Intermediate value theorem is still true for this function?

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