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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1if 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

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

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

Wallach
 2 years ago
Best ResponseYou've already chosen the best response.0sry 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
 2 years ago
Best ResponseYou've already chosen the best response.1bounded 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.

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

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

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

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

Wallach
 2 years ago
Best ResponseYou've already chosen the best response.0but 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
 2 years ago
Best ResponseYou've already chosen the best response.1the 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.

Wallach
 2 years ago
Best ResponseYou've already chosen the best response.0right 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.

Wallach
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.1let me ask this to experts!! I'll reply back later.

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

Wallach
 2 years ago
Best ResponseYou've already chosen the best response.0yes 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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