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

## 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 $|x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|<\epsilon$ so then $|x-a|<\frac{\epsilon}{|x+a||x^{2}+a^{2}|}$. Now set $|x-a|<1$. Then from the triangle inequality we know that $|x|-|a|\leq|x-a|<1$ so $|x|-|a|<1$ and it follows that $|x|<1+|a|$. Thus we have $|x-a|\leq|x|+|a|<1+2|a|$. Note that $(|x|)^{2}=|x^{2}|$ so $|x^{2}|<(1+|a|)^{2}=1+2|a|+a^{2}$. Thus we have $|x^{2}+a^{2}|<1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}$. In conclusion, we have $|x+a||x^{2}+a^{2}|<(1+2|a|)(1+2|a|+2a^{2})$ 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 $0<|x-a|<\delta\Rightarrow|x-a|<\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})} \Rightarrow|x-a||x+a||x^{2}+a^{2}|<\frac{\epsilon(1+2|a|)(1+2|a|+2a^{2})}{(1+2|a|)(1+2|a|+2a^{2})}\Rightarrow|x^{2}-a^{2}||x^{2}+a^{2}|<\epsilon\Rightarrow|x^{4}-a^{4}|<\epsilon$. This completes the proof.

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

SammyS
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## 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 $|x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|<\epsilon$ so then $|x-a|<\frac{\epsilon}{|x+a||x^{2}+a^{2}|}$. Now set $|x-a|<1$. Then from the triangle inequality we know that $|x|-|a|\leq|x-a|<1$ so $|x|-|a|<1$ and it follows that $|x|<1+|a|$. Thus we have $|x-a|\leq|x|+|a|<1+2|a|$. Note that $(|x|)^{2}=|x^{2}|$ so $|x^{2}|<(1+|a|)^{2}=1+2|a|+a^{2}$. Thus we have $|x^{2}+a^{2}|<1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}$. In conclusion, we have $|x+a||x^{2}+a^{2}|<(1+2|a|)(1+2|a|+2a^{2})$ 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 $0<|x-a|<\delta\Rightarrow|x-a|<\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})} \Rightarrow|x-a||x+a||x^{2}+a^{2}|<\frac{\epsilon(1+2|a|)(1+2|a|+2a^{2})}{(1+2|a|)(1+2|a|+2a^{2})}\Rightarrow|x^{2}-a^{2}||x^{2}+a^{2}|<\epsilon\Rightarrow|x^{4}-a^{4}|<\epsilon$. 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.

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