Proving Limit L for f(x) = x^4 with a = a

[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.

 

[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.

 

[Miike012]

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

 

[Miike012]

deleted comment.

 

[SammyS]

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.

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| .

 

[SammyS]

Hmmm, to be honest I'm stumped, I wouldn't even know where to start with
x3 + ax2 + ax +a3 to find a bound.

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| <    ?   

 

[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?

 

[SammyS]

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?

(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?

 

[Miike012]

(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?

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| < ( don't 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| + ( don't know how to relate this inequality...)

The right side is eqaul to 14|a³| + ( don't know how to relate this inequality...)

 

[SammyS]

you gave me

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

|a³| < ( don't know how to relate this inequality...)
...

The right side is equal to 14|a³| + ( don't know how to relate this inequality...)

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.)

 

[Miike012]

Yeah to be honest when I am 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?