Thread: Prove lim(x->3) x^2 = 9 by definition

1. Prove lim(x->3) x^2 = 9 by definition

Hello!

I know how to solve this task but I have certain things that I do not understand.

Solution.

$[|x^2-9|< \epsilon$ if $0 < |x-3| < \delta$

Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

$|x+3||x-3|< \epsilon$ if $0<|x-3|<\delta$

Using the triangle inequality:

$|x+3|=|(x-3)+6| \leq |x-3| + 6$

By multiplying the statement above with |x-3| (which is positive) we got:

$|x+3||x-3| \leq (|x-3| +6)|x-3|$

Now clearly $6+|x-3|< 6 + \delta$ and $|x-3|<\delta$

By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

$(6+|x-3|)|x-3|< (6 + \delta)\delta$

We can conclude that:

$|x+3||x-3| < (6 + \delta)\delta$ if $0 < |x-3| < \delta$

Now isn't $\epsilon = (6 + \delta)\delta$ ??

Why in my book they choose $(6 + \delta)\delta \leq \epsilon$ ?

And why they choose $\delta \geq 1$ and with that restriction
$(6 + \delta)\delta \leq 7\delta$ and $7\delta \leq \epsilon$.

Finally I don't understand why $\delta = min(e/7,1)$.

Could somebody possibly explain?

Thanks a lot.

2. Originally Posted by p0oint
Hello!

I know how to solve this task but I have certain things that I do not understand.

Solution.

$[|x^2-9|< \epsilon$ if $0 < |x-3| < \delta$

Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

$|x+3||x-3|< \epsilon$ if $0<|x-3|<\delta$

Using the triangle inequality:

$|x+3|=|(x-3)+6| \leq |x-3| + 6$

By multiplying the statement above with |x-3| (which is positive) we got:

$|x+3||x-3| \leq (|x-3| +6)|x-3|$

Now clearly $6+|x-3|< 6 + \delta$ and $|x-3|<\delta$

By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

$(6+|x-3|)|x-3|< (6 + \delta)\delta$

We can conclude that:

$|x+3||x-3| < (6 + \delta)\delta$ if $0 < |x-3| < \delta$

Now isn't $\epsilon = (6 + \delta)\delta$ ??

Why in my book they choose $(6 + \delta)\delta \leq \epsilon$ ?

And why they choose $\delta \geq 1$ and with that restriction
$(6 + \delta)\delta \leq 7\delta$ and $7\delta \leq \epsilon$.

Finally I don't understand why $\delta = min(e/7,1)$.

Could somebody possibly explain?

Thanks a lot.
If you reorganized your work, you will find it as follows:

$|x^2-9|=|x+3||x-3|<\epsilon$
$|x^2-9|=|x+3||x-3|<|x-3|^2+6<\epsilon$
$|x^2-9|=|x+3||x-3|<\delta^2+6<\epsilon$
$|x^2-9|=|x+3||x-3|<(\delta+6)\delta<\delta^2+6<7\delta<\epsilon$
The reason for choosing $\delta \leq 1$ is that we know with the choice, $|x^2-9|<\epsilon$ is still true. That does not mean that there is no other value of of $\delta$ besides $0<\delta\leq 1$. If $0<\delta\leq 1$ will work, why do we care about larger value of $\delta$? In other words, if the small number is closer to the limit, why bother about other large $\delta$'s that are farther from limit.

Since we know $7\delta<\epsilon$ is true, we know $\delta<\frac{\epsilon}{7}$ is also true. So we know it is safe to choose $\delta=\frac{\epsilon}{7}$, since we know with the choice, $|x^2-9|<\epsilon$ will still remain true.

If we know it will work by limiting $\delta$ to $0<\delta\leq 1$ to maintain $|X^2-9|<\epsilon$ , and we also know between 0 and 1, the is a number $\frac{\epsilon}{7}$ that also can maintain $|x^2-9|<\epsilon$ being true. Given the choices, of course we would choose $\frac{\epsilon}{7}$ over 1, since the smallest $\delta$ we could get is the one that can give us the value closest to the limit.

3. Thank you. But how you are sure that

$
(6 + \delta)\delta \leq \epsilon
$

is true?

4. Originally Posted by p0oint
Thank you. But how you are sure that

$
(6 + \delta)\delta \leq \epsilon
$

is true?
Yes, it's true.
$
(6 + \delta)\delta$
can be as large as $\epsilon$, and it can never reach beyond the value of $\epsilon$. In other words, $(6+\delta)\delta$ can only be made smaller than $\epsilon$ if you choose to make $\delta$ smaller. Because $7\delta$ is the nearest integer multiple of $\delta$, it's convinient to use 7 for dividing $\epsilon$. Suppose that $\frac{\epsilon}{(6+\delta)\delta}=1$, we can be 100% certain that when $\delta =\frac{\epsilon}{7}$, $(6+\delta)\delta <\epsilon$.

Example: For when $(6+\delta)\delta = \epsilon$, we can make table as follows:

$\delta$-------> $x=\delta+3$--------> $(6+\delta)\delta$--------> $\epsilon$

0.05 ---->3.05------------->0.3025---------->0.3025
0.50 ---->3.70------------->3.2500---------->3.2500
010 ----->4.00------------->7.0000---------->7.0000

The following are the results when $7 \delta = \epsilon$:

$\delta$-------> $x=\delta+3$--------> $(6+\delta)\delta$--------> $\epsilon$
0.05 ---->3.04------------->0.2590---------->0.3021
0.50 ---->3.42------------->2.7551---------->3.2142
010 ----->4.00------------->5.8775---------->6.8571

Now, there should be no doubt the $(6+\delta)\delta \leq \epsilon$.

5. Ok I understand now. Just one thing:

$|x-3| < \delta$

and

$\delta \leq 1$

which makes

$|x-3| \leq 1$

$2 \leq x \leq 4$

$5 \leq x + 3 \leq 7$

And we can say $|x+3| \leq 7$ and it follows $|x+3||x-3| \leq 7|x-3|$

Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta

Now we can set epsilon = 7 delta.

But one thing why in my book 7 delta <= epsilon ?

Is it because we can choose smaller values for delta then epsilon/7 ? for ex. delta= epsilon/14 so that epsilon/2 <= epsilon ?

6. Originally Posted by p0oint
Ok I understand now. Just one thing:

$|x-3| < \delta$

and

$\delta \leq 1$

which makes

$|x-3| \leq 1$

$2 \leq x \leq 4$

$5 \leq x + 3 \leq 7$

And we can say $|x+3| \leq 7$ and it follows $|x+3||x-3| \leq 7|x-3|$

Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta

Now we can set epsilon = 7 delta.

But one thing why in my book 7 delta <= epsilon ?

Is it because we can choose smaller values for delta then epsilon/7 ? for ex. delta= epsilon/14 so that epsilon/2 <= epsilon ?
For the question above it, Yes and No.

If you chose $\delta$ arbitrarily, $7\delta \leq \epsilon$ is not necessarily true, but that's not the hypothesis.

The hypothesis says, for all $\epsilon > 0$, there exist a $\delta > 0$ such that if the distance between $x$ and 3 is less than $\delta$, then $|x^2-9|<\epsilon$. In other words, you must find $\delta$ that is expressed in terms of $\epsilon$---not otherwise.

Because of the hypothesis, yes, $7\delta < \epsilon$ is true. In other words, because of the hypothesis, we can choose $\delta$ as small as we like depending on how precise we want, but we cannot choose $\delta$ such that $(6+\delta)\delta > \epsilon$.

7. Originally Posted by p0oint
Hello!

I know how to solve this task but I have certain things that I do not understand.

Solution.

$|x^2-9|< \epsilon$ if $0 < |x-3| < \delta$

Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

$|x+3||x-3|< \epsilon$ if $0<|x-3|<\delta$

Using the triangle inequality:

$|x+3|=|(x-3)+6| \leq |x-3| + 6$

By multiplying the statement above with |x-3| (which is positive) we got:

$|x+3||x-3| \leq (|x-3| +6)|x-3|$

Now clearly $6+|x-3|< 6 + \delta$ and $|x-3|<\delta$

By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

$(6+|x-3|)|x-3|< (6 + \delta)\delta$

We can conclude that:

$|x+3||x-3| < (6 + \delta)\delta$ if $0 < |x-3| < \delta$

Now isn't $\epsilon = (6 + \delta)\delta$ ??

Why in my book they choose $(6 + \delta)\delta \leq \epsilon$ ?

And why they choose $\delta \geq 1$ and with that restriction
$(6 + \delta)\delta \leq 7\delta$ and $7\delta \leq \epsilon$.

Finally I don't understand why $\delta = min(e/7,1)$.

Could somebody possibly explain?

Thanks a lot.
The following method was tought by Professor George Thomas of Massachusetts Institute of Technology:

Given: $|x^2-9|< \epsilon$ if $0 < |x-3| < \delta$

Solution:

$|x^2-9|< \epsilon$
Expanding the LHS, $|(x+3)(x-3)|< \epsilon$

Let $0<\delta < 1$, then $(x+3)$ will be in the neighborhood of 7, and

$|x-3|<1$----------------------------------eq1

Say $(x+3)=6.999$, then $|6.999(x-3)|<\epsilon$,
so that

$|x-3|<\frac{\epsilon}{7}$--------------------------eq2

Between eq1 and eq2,

$\delta = min \{\frac{\epsilon}{7}, 1\}$.

8. Thanks.

I invented one better method which does not involve arbitary numbers, but I wanted to learn other methods.

I think you made a little mistake.

It should be |x-3|<eps./7

9. Originally Posted by p0oint
Thanks.

I invented one better method which does not involve arbitary numbers, but I wanted to learn other methods.

I think you made a little mistake.

It should be |x-3|<eps./7
May I see your invention? Via Private Message if you prefer.

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prove that limit of x^2= 9 as x approaches 3

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