# Epsilon and delta

1. Jun 2, 2017

### Karol

1. The problem statement, all variables and given/known data

2. Relevant equations
Continuity:
$$\vert x-c \vert < \delta~\Rightarrow~\vert f(x)-f(c) \vert < \epsilon$$
$$\delta=\delta(c,\epsilon)$$

3. The attempt at a solution
$$\vert f(x)-f(c) \vert <\frac{1}{2}f(c)~\Rightarrow~\vert x-c \vert < \delta_1$$
So i have this δ1 but what do i do with it?
And ε=½f(c) is big, maybe it will be in the negative zone.
Maybe i have to find a δ such that $~\vert f(x)-f(c) \vert =0~$?
There is such a δ, so why was advised to take such a large ε?

2. Jun 2, 2017

### PeroK

You are still starting these proofs the wrong way round. Somehow you have to train yourself to stop writing things like:

$$\vert f(x)-f(c) \vert <\frac{1}{2}f(c)~\Rightarrow~\vert x-c \vert < \delta_1$$

You must, must, must stop yourself from doing this.

For this problem I would first try to "prove" it using a graph of the function and a geometric argument.

3. Jun 2, 2017

### PeroK

The above is continuity.

And this is what you write. You must see the difference. Every time you turn it round the wrong way.

4. Jun 2, 2017

### Karol

I was wrong at the definition of continuity:
$$\vert f(x)-f(c) \vert < \epsilon~\Rightarrow~\vert x-c \vert < \delta$$
The ε is to the intersection with x

5. Jun 2, 2017

### Staff: Mentor

No, what you have above is backwards. The implication you showed in post 1 has the implication in the right order.
In words, "If x is close to c, then f(x) will be close to f(c)"
The delta and epsilon quantify the "close to" terms.
??
In your drawing, where is x? Where is c? Is the circled point on the curve (c, f(c))?

6. Jun 2, 2017

7. Jun 2, 2017

### Dr.D

I had a friend, a fraternity brother actually, who always said that his ambition in life was to be an epsilon and delta picker.

8. Jun 2, 2017

9. Jun 3, 2017

### Karol

If f(c)>0 i can take ε small and it will still be $~\vert f(x)-f(c) \vert >0~$ and find a δ because of continuity, so why do i need the $~\epsilon=\frac{1}{2}f(c)~$?

10. Jun 3, 2017

### Staff: Mentor

In your drawing in post #8, you have $\epsilon = f(c)$, which isn't what the hint is saying.

11. Jun 3, 2017

### Karol

Thank you Mark, Dr.D and PeroK

12. Jun 3, 2017

### Staff: Mentor

@Karol, did you actually prove the theorem? The problem asks you to prove that statement, and illustrate with a sketch.

13. Jun 3, 2017

### Karol

Because of continuity i can find a δ for $~\epsilon=\frac{1}{2}f(c)~$, so in this interval: $~c-\delta<x<c+\delta~$, f(x)>0