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

  • #1
274
1

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.

[itex]f(x)=x^{2}[/itex], 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 [itex]|x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|<\epsilon[/itex] so then [itex]|x-a|<\frac{\epsilon}{|x+a||x^{2}+a^{2}|}[/itex]. Now set [itex]|x-a|<1[/itex]. Then from the triangle inequality we know that [itex]|x|-|a|\leq|x-a|<1[/itex] so [itex]|x|-|a|<1[/itex] and it follows that [itex]|x|<1+|a|[/itex]. Thus we have [itex]|x-a|\leq|x|+|a|<1+2|a|[/itex]. Note that [itex](|x|)^{2}=|x^{2}|[/itex] so [itex]|x^{2}|<(1+|a|)^{2}=1+2|a|+a^{2}[/itex]. Thus we have [itex]|x^{2}+a^{2}|<1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}[/itex]. In conclusion, we have [itex]|x+a||x^{2}+a^{2}|<(1+2|a|)(1+2|a|+2a^{2})[/itex] so then [itex]\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}[/itex]. Now for the proof.

Proof: Suppose we are given ε>0. Then choose δ=min(1,[itex]\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}[/itex]). Then [itex]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[/itex]. This completes the proof.

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

Answers and Replies

  • #2
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.

[itex]f(x)=x^{2}[/itex], arbitrary a.
That's a typo.

Should be [itex]f(x)=x^{4}[/itex] 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 [itex]|x^{4}-a^{4}|=|x^{2}-a^{2}||x^{2}+a^{2}|=|x-a||x+a||x^{2}+a^{2}|<\epsilon[/itex] so then [itex]|x-a|<\frac{\epsilon}{|x+a||x^{2}+a^{2}|}[/itex]. Now set [itex]|x-a|<1[/itex]. Then from the triangle inequality we know that [itex]|x|-|a|\leq|x-a|<1[/itex] so [itex]|x|-|a|<1[/itex] and it follows that [itex]|x|<1+|a|[/itex]. Thus we have [itex]|x-a|\leq|x|+|a|<1+2|a|[/itex]. Note that [itex](|x|)^{2}=|x^{2}|[/itex] so [itex]|x^{2}|<(1+|a|)^{2}=1+2|a|+a^{2}[/itex]. Thus we have [itex]|x^{2}+a^{2}|<1+2|a|+a^{2}+|a^{2}|=1+2|a|+2a^{2}[/itex]. In conclusion, we have [itex]|x+a||x^{2}+a^{2}|<(1+2|a|)(1+2|a|+2a^{2})[/itex] so then [itex]\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}[/itex]. Now for the proof.

Proof: Suppose we are given ε>0. Then choose δ=min(1,[itex]\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^{2})}[/itex]). Then [itex]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[/itex]. 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 [itex]|x+a||x^{2}+a^{2}|\le (1+2|a|)(1+2|a|+2a^{2})\ .[/itex]

Again, this is my personal preference.
 
  • #3
274
1
Thank you for taking your time to look over my proof. I will edit the proof and include the claim in the proof itself.
 

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