Relating $x$ and $\delta$ in Epsilon Delta proofs

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Knowing $\lvert x-b\rvert < n$.

Is there any way of having $\lvert x^k-b^k \rvert$ less than something in terms of $n$ rather than $x$?

(i.e. $\lvert x^k-b^k \rvert < \lvert n^k-b^k\rvert$)

This is part of an epsilon delta proof I am having trouble with.

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No you cannot! This fact can be re-stated as follows:

The function $f:\mathbb R \rightarrow \mathbb R$ defined by $fx)=x^k$ is not uniformly continuous for $k>1$.

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As an example, consider $k=2$. Let's say you know that $|x-a|<\epsilon$ but you don't know what $a$ is. Notice that $|x^2-a^2|=|x-a||x+a| \leq \epsilon |x+a| \leq \epsilon (2|a|+\epsilon)$. There is no way to get rid of that $|a|$ term, nor is there a way to do better estimation in order to avoid it. This is related to the fact that $f : \mathbb{R} \to \mathbb{R},f(x)=x^2$ is not uniformly continuous. An intuitive way to understand this is to notice that its derivative is not bounded. (However, there are uniformly continuous functions which are not even differentiable, so be careful!)