# Continuity of sine function

1. Jan 18, 2010

### cc10

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

Using the inequality $$|\sin(x)| < |x|$$ for $$0 < |x| < \frac{\pi}{2}$$, prove that the sine function is continuous at 0.

2. Relevant equations

Definition of continuity: A function f: R -> R is continuous at a point $$x0 \in R$$, if for any $$\epsilon > 0$$, there esists a $$\delta(\epsilon; x0)$$ such that if $$|x - x0| \leq \delta(\epsilon; x0)$$, then $$|f(x) - f(x0)| \leq \epsilon$$.

3. The attempt at a solution
To be honest, I am not really sure about this. I know that in order to prove continuity at a point, I can take the limit of the function at that point and see if it's equal to the value of the function. With this method, though, I am not using the inequality given, and therefore I cannot think of a way to do this. In class we were given no examples, so I am really lost here. I tried using the definition of continuity, but I did not manage to get anywhere that made sense.

I am sure this cannot be hard and I am probably missing something obvious. If anyone could help I'd truly appreciate it. Thank you.

2. Jan 18, 2010

### jgens

Alright, are you familiar with the squeeze theorem, because you're essentially asked to provide a formal illustration of this theorem. My suggestion is let $\delta = \varepsilon$ and then show that if $0 < |x| < \delta$ then $|\sin{(x)}| < \varepsilon$.

3. Jan 18, 2010

### cc10

I am not familiar with the squeeze theorem, but I will look into it. Thanks for the tip, I'll let you know if I get anywhere.

4. Jan 18, 2010

### LCKurtz

Hint: $|\sin x| \le |x|$.