Thread: Continuous functions help

In the following question, k =0.5.






I know how to prove the function but am unsure which part is the answer. Any help?

-1 ≤ sin(1/2x) ≤ 1, for any x ≠ 0

-xsqr ≤ xsqr sin(1/2x) ≤ xsqr, for any x ≠ 0

g(x)[= -xsqr] ≤ f(x) ≤ [xsqr =]h(x), for any x ∈ ℝ

As the functions satisfied all 3 conditions of the squeeze rule, f is continuous at 0.

p/s: Sorry, i can't seem to get the square symbol working properly.

the image is missing, you're gonna have to fix that.

Fixed. Can you see it now?

yes, it's right.

Thanks.

Anyone can help me out with this?

haha, when i said that, i tacitly told you it's correct.

Originally Posted by Krizalid

haha, when i said that, i tacitly told you it's correct.

Haha, thanks for clarifying.

The qn here is asking for the lim f(x) which i am not sure.
My lecturer said it is not 0.

Originally Posted by tottijohn

Haha, thanks for clarifying.

The qn here is asking for the lim f(x) which i am not sure.
My lecturer said it is not 0.

I think your lecturerer mispoke

The function is not only continous at 0 by also differentiable as you proved by the squeeze theorem above the limit goes to zero.

Here is a plot of the parabola's that bound the function in question.

So lim f(x) is really 0?

Thanks for the detailed explanation!
Not sure why my lecturer said it is not 0 though....