How to Prove Continuity of Sine Function at 0?

[cc10]

Homework Statement

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

Homework 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.

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.

 

[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.

 

[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.

 

[LCKurtz]

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