Homework Help: Limit question

1. Jan 14, 2012

Miike012

Determine the limit L for a given a, and prove that it is the limit.

f(x) = x^4,
a = a
L = a^4.

I get all the way to the following point...

|x-a|x+a||x^2+a^2| < ε,

I do not know how to find a "bound" for the product |x+a||x^2+a^2|. Can someone lead me in the right direction?

thank you.

2. Jan 14, 2012

Poopsilon

If you were proving uniform continuity on all of ℝ you'd have a problem (it isn't), but here you're proving point-wise continuity, which means you can define your delta in terms of more than just epsilon. Think about why this is for a bit, it was a difficult concept for me to swallow at first.

3. Jan 14, 2012

Miike012

Are you saying that my delta value is not going to be in terms of epsilon?

4. Jan 14, 2012

Miike012

deleted comment.

Last edited: Jan 14, 2012
5. Jan 15, 2012

SammyS

Staff Emeritus
Notice that (x+a)(x^2+a^2) = x3 + ax2 + ax +a3 .

So, if δ < |a|, then you can find a bound on |x+a||x^2+a^2| .

6. Jan 15, 2012

SammyS

Staff Emeritus
I had a typo in that expression. It should have been:
x3 + ax2 + a2x +a3

I also suggested letting δ < |a| .

In that case:
If |x-a| < δ

then -|a| < -δ < x-a < δ < |a|

then (after a little work) you can show that |x|< 2|a|​

This leads to |x2|< 4|a2|

and |x3|< 8|a3| .

Therefore |x3| + |ax2| +|a2x| + |a3| <    ?

7. Jan 16, 2012

Miike012

Is the answer:
....< 8|a3| + 4|a3| + 2|a3| -|a3| or

|x3| + |ax2| +|a2x| + |a3| - |a3| < 13|a3|...

is this correct? If not am I close?

8. Jan 16, 2012

SammyS

Staff Emeritus
(Use the X2 button on the "Go Advanced" message box to display exponents. Otherwise use LaTeX.)

13|a3| is not right.

How did you get that?

9. Jan 16, 2012

Miike012

you gave me

|x| < 2|a|
|x2| < 4|a2| ,
|x3| < 8|a3|, and
|x3| + |ax2| + |a2x| + |a3| < .

so
|x3| < 8|a3|,
|ax2| < 4|a*a2| ,
|a2*x| < 2|a2*a|

|a3| < ( dont know how to relate this inequality...)

Then I added the above 4 lines together and got,

|x3| + |ax2| +|a2*x| + |a3| < 8|a3| + 4|a*a2| + 2|a2*a| + ( dont know how to relate this inequality...)

The right side is eqaul to 14|a3| + ( dont know how to relate this inequality...)

10. Jan 16, 2012

SammyS

Staff Emeritus
The way to deal with |a3| is simply that |a3| = |a3|

So if
|x| < 2|a|,
|x2| < 4|a2| ,
|x3| < 8|a3|,
and
|a3| is what it is:​
then
|x3| + |ax2| + |a2x| + |a3| < 8|a3| + 4|a2|∙|a| + 2|a|∙|a2| + |a3|​

This will work fine as long as a ≠ 0.

The case in which a = 0 can be handled pretty easily, after you have conquered the more general case. (It is also true that you can work out a method that handles all values of a, but the algebra will be tougher than that above and that looks as if it has given you enough of a problem.)

11. Jan 16, 2012

Miike012

Yeah to be honest when Im working with inequalities I am totally lost... Is there any place I can read up on inequalities so maybe I can understand them better and learn how to manipulate them better?

and what is the more "general" way to do it?

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