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

1. Sep 20, 2010

### Robultronic

1. The problem statement, all variables and given/known data

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 [-$$\sqrt{\pi}$$, $$\sqrt{\pi}$$ ]

2. Relevant equations

3. 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.

2. Sep 20, 2010

### vela

Staff Emeritus
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?

3. Sep 20, 2010

### 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.

4. Sep 20, 2010

### vela

Staff Emeritus
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.

5. Sep 20, 2010

### 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?

6. Sep 21, 2010

### 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.