Continuity: Epsilon & Delta Homework

[Karol]

Homework Statement

Homework Equations

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

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?

 

[PeroK]

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?

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.

 

[PeroK]

Homework Equations

Continuity:
$$\vert x-c \vert < \delta~\Rightarrow~\vert f(x)-f(c) \vert < \epsilon$$

The above is continuity.

The Attempt at a Solution

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

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

 

[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

 

[Mark44]

I was wrong at the definition of continuity:
$$\vert f(x)-f(c) \vert < \epsilon~\Rightarrow~\vert x-c \vert < \delta$$

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.

The ε is to the intersection with x

??

View attachment 204730

In your drawing, where is x? Where is c? Is the circled point on the curve (c, f(c))?

 

[PeroK]

I was wrong at the definition of continuity:
$$\vert f(x)-f(c) \vert < \epsilon~\Rightarrow~\vert x-c \vert < \delta$$

 

[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.

 

[Karol]

 

[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 $~\epsilon=\frac{1}{2}f(c)~$?

 

[Mark44]

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

 

[Karol]

Thank you Mark, Dr.D and PeroK

 

[Mark44]

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

 

[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