Prove limit of x^5 = a^5 using epsilon delta

$\displaystyle \lim_{x\to a}x^5=a^5 $

I understand that the power rule for limits could be used here but my teacher wants it done using the epsilon delta definition.

1st attempt (though I think it's obviously wrong since my $\displaystyle \delta$ could be undefined):

$\displaystyle |x-a|< \delta \rightarrow |x^5-a^5|< \epsilon$

$\displaystyle |x^5-a^5|=|(x-a)(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|< \epsilon$

$\displaystyle |x-a|< \frac{\epsilon}{|(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|}$

$\displaystyle Choose: \delta = \frac{\epsilon}{|(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|}$

Re: Prove limit of x^5 = a^5 using epsilon delta

You're right that you need to factor. But, you shouldn't have x terms in your choice of delta. Try getting a bound on the big factor (x^4+...) by making a bound on x.

Re: Prove limit of x^5 = a^5 using epsilon delta

Do you mean something very similar to the below example?

Are you suggesting the bound I make is something like $\displaystyle |x-a|<1$?

which implies $\displaystyle -1<x-a<1$

$\displaystyle a-1<x<a+1$

$\displaystyle x^4<(a+1)^4$

$\displaystyle |x^3|<|(a+1)^3|$

http://oi39.tinypic.com/ambd6c.jpg

Re: Prove limit of x^5 = a^5 using epsilon delta

Indeed. But don't forget that you want an upper bound, and you certainly have that x< |a|+1 So, use this to get aqn upper bound on the big polynomial factor, take delta to be the minimum of 1 and epsilon/(bound on poly)