Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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??

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Perhaps it's for all x in its domain?
the function is not continuous at x=0, are you missing something?
it's for \[f(x) \rightarrow R:R\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
I might be missing some thing but I'm not sure what I would have uploaded the question but it's in Hebrew
that I know. that's because x while x-->0 equals 0 and sin(1/x) is a Bounded function
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
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.
that's what I thought
f(x) = x sin(1/x) is particularly interesting ... being continuous but not differentiable at x=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 \[\lim_{x \rightarrow 0} f(x)*g(x) = 0\]
which is the case that you're presenting
yeah ... the case is different from what you are asking.
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
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.
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.
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 :)
let me ask this to experts!! I'll reply back later.
tnk u
http://chat.stackexchange.com/transcript/message/6829811#6829811
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question