Question

Epsilon Delta with infinity?

Answer 1

x->infinity 1/x^4=0

Answer 2

In this case, you'd say \(x>\delta\) rather than \(|x-a|<\delta\)

Answer 3

Everything else stays the same.

Answer 4

\[\forall\epsilon>0 \space \exists M>0 \space s.t. x>m \implies|f(x)-L|<\epsilon\]

Answer 5

we may assume m is positive, but it does not have to be....

Answer 6

so
\[let \space m_0=\frac{1}{\sqrt[4]{\epsilon}}\]\[let \space m =max\{1,m_0\}\]then
\[x>m \implies|\frac{1}{x^4}-0|=\frac{1}{x^4}<\frac{1}{m^4}<\epsilon\]

Answer 7

that shuold be\[\frac{1}{x^4}\le\frac{1}{m^4}<\epsilon\]\

Answer 8

i think this is correct.... how does that look @wio ?

Answer 9

How do you know \(x^{-4}\leq m^{-4}\)?

Answer 10

It looks right though

Answer 11

because x>m and m>=1
so x,m >=1 and m<x
so 1/x^4<1/m^4