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Wallach
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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??
 one year ago
 one year 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??
 one year ago
 one year 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?
 one year ago

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

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

Wallach Group TitleBest ResponseYou've already chosen the best response.0
that's what I thought
 one year 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.
 one year 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 \[\lim_{x \rightarrow 0} f(x)*g(x) = 0\]
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... the case is different from what you are asking.
 one year 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
 one year 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.
 one year 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.
 one year 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 :)
 one year ago

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

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