Question

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??

Answer 1

Perhaps it's for all x in its domain?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

that's what I thought

Answer 10

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

Answer 11

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

Answer 12

which is the case that you're presenting

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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.

Answer 17

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 :)

Answer 18

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

Answer 19

tnk u

Answer 20

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

Answer 21

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?