Homework Help: 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: $|\sin x| \le |x|$.