Show that f(x) = sin(x^2) is continuous for all a [-sqrt(pi), sqrt(pi).

[Robultronic]

Homework Statement

Given f : [-\sqrt{\pi}, \sqrt{\pi} ] \rightarrow [-1, 1]

f(x) = sin(x^{2})

a)
Show that f is continuous for all a \in [-\sqrt{\pi}, \sqrt{\pi} ]

b)
Find a \delta so that |x - y| \leq \delta implies that
|f(x) - f(y)| \leq 0.1 for all x and y in [-\sqrt{\pi}, \sqrt{\pi} ]

Homework Equations

The Attempt at a Solution

I honestly don't know where to begin. But would be very pleased for any help. A full solution would be the optimal though. But anything helps.

 

[vela]

If you want to go back to the definition of continuity, you need to show that for every point a in the interval,

\lim_{x \to a} f(x) = f(a)

(with the appropriate one-sided limit at the endpoints). If you can assume other facts, then you can argue f(x) is continuous in a less tedious way. What do you think you're supposed to do?

 

[Robultronic]

I'm supposed to use the definition of continuity.

I know how to show continuity at a point, but have no clue of how to do it for an interval.

 

[vela]

You just have to show f is continuous at every point in the interval. It essentially amounts to saying "Let a∈[-√π,√π]" and showing f is continuous at x=a like usual.

 

[Robultronic]

Lol, that simple, thx. I kept thinking I had to use the interval in the proof somehow and not just say a = the interval. OK, that I can do. How about b?

 

[Robultronic]

I think I solved it, I will write it down when I get home so someone can tell if its correct or not, but have to go now.