Finding Delta for Epsilon in x^3 Function for c=5

[Karol]

Homework Statement

find a δ for a given ε for f(x)=x³ around c=5:
$$\vert x-5\vert<\delta~\Rightarrow~\vert x^3-5^3 \vert < \epsilon$$

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

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

 

[PeroK]

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

 

[BvU]

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

 

[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?

 

[PeroK]

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?

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?

 

[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 $~\delta<\frac{\epsilon}{91}$

 

[PeroK]

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 $~\delta<\frac{\epsilon}{91}$

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 $|x-5| < \dots$

 

[Karol]

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

 

[BvU]

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

 

[PeroK]

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

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.

 

[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

 

[PeroK]

I wasn't told, in the original question, that $~|x-5| < 1~$

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

 

[PeroK]

I wasn't told, in the original question, that $~|x-5| < 1~$/QUOTE]

You weren't told that ##|x -5| < \epsilon

 

[PeroK]

I wasn't told, in the original question, that $~|x-5| < 1~$

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

 

[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

 

[PeroK]

And even if i solve correctly for $~|x - 5| < 1~$ it's not a general answer for all x

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.

 

[Karol]

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?

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)

 

[PeroK]

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)

Unfortunately, then, I don't think I can help you any further.

 

[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

 

[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 $\varepsilon$

 

[PeroK]

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

I think you probably need some one-to-one tuition on this, as your questions are difficult to answer.

I don't think @BvU is nit-picking. You need to say something about large $\epsilon$. And, it's critical to establish an intermediate bound on $|x-5|$ in order to get the main, variable bound based on $\epsilon$.

More fundamentally, you need to look carefully at why we can take $|x-5|<1$. It helps to limit $\delta$, even when $\epsilon/91$ is large.

 

[PeroK]

@Karol

Let me show you a rigorous proof for $x^2$ at $x = 5$. Note that you can find this online in any case. E.g.

https://www.freemathhelp.com/forum/threads/62586-rigorously-prove-x-2-is-continuous

But, my version will be a little more step-by-step.

Let $\epsilon > 0$.

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

Let $|x-5| < 1$, then:

$4 < x < 6$, $|x+5| < 11$ and:

$|x^2 - 5^2| < 11|x-5|$

Now, further, let $|x-5| < \epsilon/11$. Then:

$|x^2 - 5^2| < \epsilon$

Putting this all together we have shown that:

$\forall \ \epsilon > 0, |x-5| < min \{ 1, \frac{\epsilon}{11} \} \ \Rightarrow \ |x^2 - 5^2| < \epsilon$

Thus, we have proved that $x^2$ is continuous at $x=5$.

(Note that I didn't in fact use the symbol $\delta$. If this offends or confuses you, then you can write:

Let $\delta = min \{ 1, \frac{\epsilon}{11} \}$, then:

$\forall \ \epsilon > 0, |x-5| < \delta \ \Rightarrow \ |x^2 - 5^2| < \epsilon$)

Try to work through this and understand the sequence of steps and the logic.

 

[Karol]

Thank you very much PeroK and BvU