Understanding a problem that uses episolon delta defintion

[Genericcoder]

let E = episolon and D = delta;

the problem is as follows:

let f(x) = (2x^2 - 3x + 3). prove that lim as x approaches 3 f(x) = 21,


we write |f(x) - 21| = |x^2 + 2x - 15| = |x + 5||x - 3|


to make this small, we need a bound on the size of |x + 5| when x is close to 3. For example,
if we arbitarily require that |x - 3| < 1 then


|x + 5| = |x - 3 + 8| <= |x - 3| + |8| < 1 + 8 = 9

to make E f(x) within E of 21, we want to have |x + 5| < 9 and |x - 3| < E/9

I don't understand how did he get E/9 |x - 3| < E/9 ?

 

[jbunniii]

let E = episolon and D = delta;

the problem is as follows:

let f(x) = (2x^2 - 3x + 3). prove that lim as x approaches 3 f(x) = 21,

This is clearly false. Note that $f(3) = 12$, not $21$. And $f$ is continuous, so $\lim_{x \rightarrow 3}f(x) = f(3)$. Is there a typo?

 

[jbunniii]

Also, how did you get this:

we write |f(x) - 21| = |x^2 + 2x - 15|

from $f(x) = 2x^2 - 3x + 3$? Are you sure you are not mixing up two different problems?

 

[Genericcoder]

it should be 2x^2 + 2x + 6 your right ! but the same logic holds for the problem that I typed I don't know how did he get |x - 3| < E/9...

 

[Genericcoder]

srry I had a type your right it should be 2x^2 + 2x + 6 !

 

[jbunniii]

it should be 2x^2 + 2x + 6 your right !

That still doesn't have a limit of $21$ as $x \rightarrow 3$. It's hard to help if you don't write down the correct problem!

From this line:

|f(x) - 21| = |x^2 + 2x - 15| = |x + 5||x - 3|

I am going to assume that you meant $f(x) = x^2 + 2x + 6$, which does have the limit $21$ as $x \rightarrow 3$. So, proceeding from that assumption:

Clearly it's not going to be a problem to make $|x-3|$ as small as we like as $x \rightarrow 3$. So as your narrative says, we just need to make sure that $|x+5|$ doesn't grow without bound as we shrink $|x-3|$ to zero.

I assume you are OK with the logic that shows that if $|x-3| < 1$, then $|x+5| < 9$.

So now our goal is to make $|x+5||x-3| < \epsilon$. We already know we need $|x-3|< 1$ in order for the bound $|x+5| < 9$ to be valid. If we ALSO had $|x-3| < \epsilon / 9$, then we could conclude that
$$|x+5||x-3| < 9 \cdot \frac{\epsilon}{9} = \epsilon$$
So how should we define $\delta$?

 

[Genericcoder]

but why did u assume |x - 3| < E/9 out of nowhere?

 

[jbunniii]

but why did u assume |x - 3| < E/9 out of nowhere?

Because I recognized that was the factor I needed in order to get $|x+5||x-3| < \epsilon$, given that $|x+5| < 9$.

So all that needs to be done is to show that the assumption can be achieved. In other words, we need a $\delta$ such that if $|x-3| < \delta$, then both of the assumptions that we have made are satisfied, namely $|x-3| < \epsilon/9$ and $|x-3| < 1$. How can I choose $\delta$ to guarantee this?

 

[Genericcoder]

oh I see oke so in order to achieve this we make E = min{1,E/9} right?

 

[jbunniii]

oh I see oke so in order to achieve this we make E = min{1,E/9} right?

I assume you mean D = min{1,E/9}. That is correct.

 

[Genericcoder]

can u give me a website that has a lot of examples on epsilon delta proof of limits?

 

[jbunniii]

Have you looked in the "Mathematics Learning Materials" section?

https://www.physicsforums.com/forumdisplay.php?f=178