You are Here: Home >< Maths

# epsilon delta q watch

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

2. (Original post by cooldudeman)
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?

3. (Original post by Shillington)
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
4. (Original post by cooldudeman)
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
5. (Original post by Shillington)
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
Attached Images

6. (Original post by cooldudeman)
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)?
7. (Original post by Shillington)
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
Attached Images

8. (Original post by cooldudeman)
this is the original q. 1st

Posted from TSR Mobile
See here for methods of showing |sin(x)-sin(y)|<=|x-y|
http://math.stackexchange.com/questi...us-on-mathbb-r

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 17, 2013
Today on TSR

### Cambridge interviews

Find out which colleges are sending invitations

### University open days

• University of East Anglia
Fri, 23 Nov '18
• Norwich University of the Arts
Fri, 23 Nov '18
• Edge Hill University
Sat, 24 Nov '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE