epsilon delta q

does it make sense, what I did in the pic? I considered the Max value of sinx-sinx0.
Posted from TSR Mobile

It is unnecessary to use this delta=min(epsilon,2). Can you see why taking delta=epsilon will work? The "work" in this question is showing that |sin(x)-sin(x_0)|<=|x-x_0|, where did you show this?

am I meant to say epsilon is greater than 2?
|sin(x)-sin(x_0)|<=2
2< |x-x_0|<epsilon but I get the feeling that this doesn't make sense


Posted from TSR Mobile

I don't quite get what you're asking. You don't say anything about epislon, it is given to you. You must then show that when delta is appropriately chosen, the resulting sequence of inequalities holds i.e.

|x-x_0|<delta
=>
|sin(x)-sin(x_0)|<epsilon

this is all I can think of to show
|sin(x)-sin(x_0)|<=|x-x_0|?
Posted from TSR Mobile

In all likelihood it was a given in the question, have you got the question to hand?

I can't think of a neat way of showing it otherwise, have you even come across the definition of sin(x)?

this is the original q. 1st

Posted from TSR Mobile