# Epsilon and delta of x^3

1. May 30, 2017

### Karol

1. The problem statement, all variables and given/known data
find a δ for a given ε for f(x)=x3 around c=5:
$$\vert x-5\vert<\delta~\Rightarrow~\vert x^3-5^3 \vert < \epsilon$$

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
$$\vert x^3-5^3 \vert =\vert x-5 \vert \cdot \vert x^2+5x+25 \vert$$
$$=\vert x-5 \vert \cdot \vert (x-5)^2+15(x-5)+75 \vert < \epsilon$$
$$\delta \cdot (\delta^2+15\delta+75)<\epsilon$$

2. May 30, 2017

### PeroK

Hint: look at $x^2+5x+25$. What can you do with that if $x$ is "close enough" to $5$?

3. May 30, 2017

### BvU

B) If I give you a value for epsilon, can you give me a value for delta (a value for delta itself, that is) ?

4. May 30, 2017

### Karol

If x is close to 5 from the left side then:
$$\vert x^3-5^3 \vert =\vert x-5 \vert \cdot \vert x^2+5x+25 \vert < 75\delta~\rightarrow~\delta<\frac{\epsilon}{75}$$
But if x is on the right?

5. May 30, 2017

### PeroK

I was thinking more that if $x$ is close to $5$ then $x^2 + 5x + 25 \approx 75$.

And, if $|x - 5| < 1$, say, then how big can $x^2 + 5x + 25$ be?

6. May 30, 2017

### Karol

if $~|x-5|<1$:
$$\vert x^3-5^3 \vert \triangleq\epsilon=\vert x-5 \vert \cdot \vert x^2+5x+25 \vert < 91\delta~\rightarrow~\delta>\frac{\epsilon}{91}$$
But it should be $~\delta<\frac{\epsilon}{91}$

7. May 30, 2017

### PeroK

I think you are getting yourself confused in some way. You start with:

$|x^3 - 5^3| = |x-5||x^2 +5x + 25|$

Now you let $|x -5| < 1$, which implies $4 < x < 6$, which implies that $|x^2 +5x + 25| < 91$

Putting that together you have:

$|x -5| < 1 \ \Rightarrow \ |x^3 - 5^3| < 91|x-5|$

Now, if you let $\epsilon > 0$ can you finish it off?

Hint: Let $|x-5| < \dots$

8. May 31, 2017

### Karol

$$\delta\triangleq |x-5|=\frac{|x^3-5^3|}{|x^2+5x+25|}=\frac{\epsilon}{|x^2+5x+25|}<\frac{\epsilon}{91}$$
$$\Rightarrow\delta<\frac{\epsilon}{91}$$

9. May 31, 2017

### BvU

I give you $\epsilon = 200\$. Then $\delta =2 < {200\over 91}\$ is not good enough ...

10. May 31, 2017

### PeroK

That's the background analysis, but it's not a proof, because you've got no explanation of what you are doing and the implication is the wrong way at the end.

What you've shown is:

$|x - 5| = \delta \ \Rightarrow \ \delta < \frac{\epsilon}{91}$

(I'm sorry to say I don't know what the little triangle means.)

You need to turn the logic round, so that for a given $\epsilon$ you have a well-defined $\delta$.

Also, don't forget that you needed $|x-5| < 1$ as well.

11. May 31, 2017

### Karol

I wasn't told, in the original question, that $~|x-5| < 1~$.
And for |x-5|<1 it's correct $~\delta < \frac{\epsilon}{91}$. but for all real numbers on x i don't know what to do since |x-5| doesn't appear in (x2+5x+25)
$\triangleq$ means definition, a symbol given to

12. May 31, 2017

### PeroK

You weren't told that $|x -5| < \epsilon /91$ either. So, where did that come from?

13. May 31, 2017

### PeroK

14. May 31, 2017

### PeroK

You weren't told that $|x -5| < \epsilon /91$ either.

15. May 31, 2017

### Karol

$|x -5| < \epsilon /91~$ came from the condition $~|x - 5| < 1$ you proposed, and i guess $~\delta<\frac{\epsilon}{91}~$ satisfies this condition, but i am not sure.
And even if i solve correctly for $~|x - 5| < 1~$ it's not a general answer for all x

16. May 31, 2017

### PeroK

Yes it is a solution. You are free to choose $x$ as close to 5 as you like. That's the whole point. If you can't choose $x$ close to $5$ how can you prove continuity?

If you look at $x = 1,000,000$ then even large $\epsilon$ is a problem.

17. May 31, 2017

### Karol

I don't think it's a solution, since the definition of continuity first chooses ε and derives a δ. i choose ε as small as i want and the δ becomes small, x comes close to 5.
Yes, large ε is out of the domain |x-5|<1 (ε is not in the domain, it's on the Y axis, of course)

18. May 31, 2017

### PeroK

19. Jun 1, 2017

### Karol

I am sorry i sounded not polite, PeroK, please help, i have no idea how to solve even for |x-5|<1, as you suggested and i am very frustrated

20. Jun 1, 2017

### BvU

You have shown that for $\varepsilon < 91 \$ the condition $\delta < {\varepsilon \over 91} \$ is sufficient. I think that should be good enough for this exercise.

Personally, all I wanted to point out was that: just $\delta < {\varepsilon \over 91} \$ in itself is not really sufficient if you take the exercise wording literally.

In analysis statements are made like 'For all $\varepsilon > 0 \$ there exists a $\delta > 0\$ such that ... ' with the silent implication/intention that $\varepsilon$ can be made as small as desired (you $\$ look at/worry about $\$ the possibility to find a $\delta$ for $\varepsilon \downarrow 0$ ).

Outside of such a context nitpckers like me can point out that a simple expression for $\delta$ may fail for big $\varepsilon$