Proving Limit of x^4 as x→p is p^4

[asdf60]

how would i prove that lim of x^4 as x->p is p^4? x^4-p^4 = (x-p)(x+p)(x^2+p^2). I'm having trouble controlling the (x+p)(x^2+p^2) term without having to resort to proving for p > 0 and p < 0 separately.

 

[shmoe]

A standard idea is to place an upper bound on your delta to restrict x, then use this to bound the (x+p)(x^2+p^2) part. If you knew delta<=|p|, can you find an upper bound for |(x+p)(x^2+p^2)|?

This is assuming p is not 0. You could modify the delta above to something that would work in this case, or just deal with it separately.

 

[asdf60]

I get |x+p| < |3p| and |x^2+p^2| < 5p^2 if delta < |p|. So then 15p^3 is an upper bound. Is this correct? And as far as dealing with p = 0, can I just add one to the denominator? How did you come up with using |p| for delta?

 

[shmoe]

That looks good for an upper bound. If p=0, you can't take |p| as an upper bound for delta, you'd want something like |p|+1, but this makes everything else slightly uglier (but perfectly doable). Taking |p| as an upper bound for delta was pretty arbitrary, it just made sure x was positive (or negative if p was negative) and this looked a little cleaner. You could just as well take 1 as an upper bound for delta here.

In general you want to make sure your delta is small enough that the piece you're trying to bound is actually bounded on |x-p|<delta, usually you're trying to avoid zeros in the denominator or where the function otherwise "blows up"

 

[asdf60]

Thanks a lot for the help. I guess I had it right before, but it was really a mess without using |p| as the delta upper. as you said, this makes the upper bound much cleaner. For p=0, i just used delta = (epsilon)^(1/4).