Epsilon delta proof that x^4 goes to a^4 as x goes to a

[DeadOriginal]

Homework Statement

Determine the limit l for a given a and prove that it is the limit by showing how to find δ such that |f(x)-l|<ε for all x satisfying 0<|x-a|<δ.

f(x)=x^{2}, arbitrary a.

Homework Equations

I will incorporate the triangle inequality in this proof.

The Attempt at a Solution

We want to be able to find a δ such that if 0<|x-a|<δ then |x^4-a^4|<ε. Working backwards we see that |x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|&lt;\epsilon so then |x-a|&lt;\frac{\epsilon}{|x+a||x^{2}+a^{2}|}. Now set |x-a|&lt;1. Then from the triangle inequality we know that |x|-|a|\leq|x-a|&lt;1 so |x|-|a|&lt;1 and it follows that |x|&lt;1+|a|. Thus we have |x-a|\leq|x|+|a|&lt;1+2|a|. Note that (|x|)^{2}=|x^{2}| so |x^{2}|&lt;(1+|a|)^{2}=1+2|a|+a^{2}. Thus we have |x^{2}+a^{2}|&lt;1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}. In conclusion, we have |x+a||x^{2}+a^{2}|&lt;(1+2|a|)(1+2|a|+2a^{2}) so then \frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}. Now for the proof.

Proof: Suppose we are given ε>0. Then choose δ=min(1,\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}). Then 0&lt;|x-a|&lt;\delta\Rightarrow|x-a|&lt;\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})} \Rightarrow|x-a||x+a||x^{2}+a^{2}|&lt;\frac{\epsilon(1+2|a|)(1+2|a|+2a^{2})}{(1+2|a|)(1+2|a|+2a^{2})}\Rightarrow|x^{2}-a^{2}||x^{2}+a^{2}|&lt;\epsilon\Rightarrow|x^{4}-a^{4}|&lt;\epsilon. This completes the proof.

Could someone take a look at this for me? I feel like something isn't right with it.

 

[SammyS]

Homework Statement

Determine the limit l for a given a and prove that it is the limit by showing how to find δ such that |f(x)-l|<ε for all x satisfying 0<|x-a|<δ.

f(x)=x^{2}, arbitrary a.

That's a typo.

Should be f(x)=x^{4} of course.

Homework Equations

I will incorporate the triangle inequality in this proof.

The Attempt at a Solution

We want to be able to find a δ such that if 0<|x-a|<δ then |x^4-a^4|<ε. Working backwards we see that |x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|&lt;\epsilon so then |x-a|&lt;\frac{\epsilon}{|x+a||x^{2}+a^{2}|}. Now set |x-a|&lt;1. Then from the triangle inequality we know that |x|-|a|\leq|x-a|&lt;1 so |x|-|a|&lt;1 and it follows that |x|&lt;1+|a|. Thus we have |x-a|\leq|x|+|a|&lt;1+2|a|. Note that (|x|)^{2}=|x^{2}| so |x^{2}|&lt;(1+|a|)^{2}=1+2|a|+a^{2}. Thus we have |x^{2}+a^{2}|&lt;1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}. In conclusion, we have |x+a||x^{2}+a^{2}|&lt;(1+2|a|)(1+2|a|+2a^{2}) so then \frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}. Now for the proof.

Proof: Suppose we are given ε>0. Then choose δ=min(1,\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}). Then 0&lt;|x-a|&lt;\delta\Rightarrow|x-a|&lt;\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})} \Rightarrow|x-a||x+a||x^{2}+a^{2}|&lt;\frac{\epsilon(1+2|a|)(1+2|a|+2a^{2})}{(1+2|a|)(1+2|a|+2a^{2})}\Rightarrow|x^{2}-a^{2}||x^{2}+a^{2}|&lt;\epsilon\Rightarrow|x^{4}-a^{4}|&lt;\epsilon. This completes the proof.

Could someone take a look at this for me? I feel like something isn't right with it.

I suppose it depends who grades the proof.

You have done a thorough job explaining how you get δ from ε. But that's not part of your formal proof.

Personally, I like to see everything that's needed for the proof, in the proof.

So in the proof itself, I like to see you support your claim that |x+a||x^{2}+a^{2}|\le (1+2|a|)(1+2|a|+2a^{2})\ .

Again, this is my personal preference.

 

[DeadOriginal]

Thank you for taking your time to look over my proof. I will edit the proof and include the claim in the proof itself.