# Prove limits using epsilon delta definition

1. Nov 1, 2012

### fire sword

1. The problem statement, all variables and given/known data
http://store2.up-00.com/Sep12/JB498124.jpg [Broken]

2. The attempt at a solution
No attempts cuz i can't understand how to solve it

Last edited by a moderator: May 6, 2017
2. Nov 1, 2012

### arildno

HOW have you been given definitions of the trig functions??

3. Nov 1, 2012

### fire sword

the teacher want it by epsilon delta def only

4. Nov 1, 2012

### fire sword

if u have examples for these types of problems u could provide me with it
and i will try to understand

5. Nov 1, 2012

### arildno

There is no epsilon definition of the sine and cosine functions.
I asked you how you have been defined THOSE functions?
Have they been proven, or defined as, continuous, for example?
Have you been given them as by representations of solutions of eigen value problems, or as exaples of power series?

6. Nov 1, 2012

### fire sword

he only explained delta-epsilon definition for normal functions and gave us these to think how to prove them

7. Nov 1, 2012

### arildno

Well, then ASSUME that he regards that sine and cosine are CONTINUOUS functions (and that that is, therefore, allowable for you, as a student).
Make this an EXPLICIT assumption if you make a detailed reply.
Which of the limits are you thereby able to prove, given that assumption?

8. Nov 1, 2012

### micromass

Well, we really can't help you until you give us a definition of sine and cosine...

9. Nov 1, 2012

### Zondrina

You could do this without any definitions for sin and cos.

Are you allowed to use the mean value theorem yet is my only question? If so then these are all possible to do using ε-δ.

Otherwise the squeeze theorem would be the only other plausible way.

10. Nov 1, 2012

### micromass

How can you possibly show that sin is continuous without defining sin?

11. Nov 1, 2012

### Zondrina

I suppose I see what you're saying here ^, but continuity of sin alone over ℝ would be sufficient enough to show that |sinx - sina| < ε

I would assume the continuity of sin was trivial though?

12. Nov 1, 2012

### micromass

The first question requires him to show continuity of sine. So I don't think he can assume that.

13. Nov 1, 2012

### Zondrina

Hmm that is a problem then. If you're not even given the fact that sin is continuous or any info about sin ( let alone the mean value theorem which is much later in any calculus course ), I don't think it would be possible to show this...

According to the OP, the professor threw a curve ball. More info would be needed I guess.

14. Nov 1, 2012

### arildno

Nope.
It might be the question-for-the-Dummies:
Merely REFER to the general definition of continuity in order to show that you understand what it actually means that a continuous function is..continuous.

Given the next questions, though, OP has simply forgotten what he has told in lecture about the BEHAVIOUR of the trig functions.

15. Nov 1, 2012

### Arian.D

I hope that I'm not mistaken, but just start with the fact that $|sin(x)|\leq|x|$ for all real numbers.

then prove that: $|sin(x)-sin(a)| \leq |x-a|$
you can also prove that $|cos(x)-cos(a)| \leq |x-a|$

That tells you why sin(x) and cos(x) is continuous. This solves the first and second problems.

But it's still very hard to prove all of those limits with epsilon-delta definition. also for the last two ones you can't write an epsilon-delta definition, because the limit tends to go to infinity.

16. Nov 1, 2012

### jbunniii

If you define $\cos(\theta)$ and $\sin(\theta)$ in the usual geometric way, as the x- and y- coordinates of a point at angle $\theta$ on the unit circle, and you're willing to accept a simple geometric argument that these coordinates must approach (1,0) as $\theta$ approaches 0, then this means that $\cos$ and $\sin$ are continuous at 0. From that, you can use trigonometric identities to get continuity elsewhere, for example
$$\sin(x+h) - \sin(x) = \sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)$$
so
\begin{align*} \lim_{h \rightarrow 0} (\sin(x+h) - \sin(x)) &= \sin(x) \left(\lim_{h \rightarrow 0}\cos(h)\right) + \cos(x)\left(\lim_{h \rightarrow 0}\sin(h)\right) - \sin(x) \\ &= \sin(x) \cos(0) + \cos(x) \sin(0) - \sin(x) \\ &= \sin(x) + 0 - \sin(x) = 0 \end{align*}
and therefore $\sin$ is continuous at $x$. I'm not sure how you would turn this into a fully rigorous $\epsilon-\delta$ argument, however.