## Wallach 3 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??

1. terenzreignz

Perhaps it's for all x in its domain?

2. experimentX

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

3. Wallach

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

4. 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

5. Wallach

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

6. Wallach

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

7. Wallach

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

8. 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.

9. Wallach

that's what I thought

10. experimentX

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

11. Wallach

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 $\lim_{x \rightarrow 0} f(x)*g(x) = 0$

12. Wallach

which is the case that you're presenting

13. experimentX

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

14. Wallach

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

15. 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.

16. Wallach

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.

17. Wallach

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 :)

18. experimentX

19. Wallach

tnk u

20. experimentX
21. Wallach

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?