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Wallach
Group Title
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??
 2 years ago
 2 years ago
Wallach Group Title
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??
 2 years ago
 2 years ago

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terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Perhaps it's for all x in its domain?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the function is not continuous at x=0, are you missing something?
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
it's for \[f(x) \rightarrow R:R\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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/isexpleft1overx2rightdifferentiableatx0
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
I might be missing some thing but I'm not sure what I would have uploaded the question but it's in Hebrew
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
that I know. that's because x while x>0 equals 0 and sin(1/x) is a Bounded function
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
that's what I thought
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
f(x) = x sin(1/x) is particularly interesting ... being continuous but not differentiable at x=0.
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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 \[\lim_{x \rightarrow 0} f(x)*g(x) = 0\]
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
which is the case that you're presenting
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... the case is different from what you are asking.
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let me ask this to experts!! I'll reply back later.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
http://chat.stackexchange.com/transcript/message/6829811#6829811
 2 years ago

Wallach Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago
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