Show that $f(x) = x^{p}$ is not uniformly continuous on $\mathbb{R}$ if $p > 1$
Attempt:
Consider the sequences $$x_{n} = n + \frac{1}{n} \\ y_{n} = n$$
Examining these sequences we have:
$$|x_n - y_n| = |n + \frac{1}{n} - n| = \Bigg|\frac{1}{n}\Bigg| \\ \Rightarrow \lim_{n \rightarrow \infty} \Bigg|\frac{1}{n} \Bigg| = 0$$
But:
$$|f(x_{n}) - f(y_{n})| = \Bigg| (n + \frac{1}{n})^{p} - n^{p} \Bigg| = \Bigg|\sum_{k = 0}^{p}\binom{p}{k}n^{p-k}\Bigg(\frac{1}{n} \Bigg)^{k} - n^{p} \Bigg| \\ \Rightarrow \Bigg|\sum_{k = 1}^{p}\binom{p}{k}n^{p-k}\Bigg(\frac{1}{n} \Bigg)^{k} \Bigg| > p \ - used \ binomial \ thm \ and\ just\ subtracted\ the\ first\ term. $$
Therefore $f(x) = x^{p}$ is not uniformly continuous.
Thoughts:
I think I over complicated things by wanting to use $$\Bigg|\sum_{k = 1}^{p}\binom{p}{k}n^{p-k}\Bigg(\frac{1}{n} \Bigg)^{k} \Bigg| > p$$
In fact I probably have to prove such a claim, which I had difficulty doing by induction. I essentially just replicated the proof that would usually be used to show $f(x) = x^2$ is not uniformly continuous over all of $\mathbb{R}$.If this is the right idea, what would be a way to establish that last inequality from the binomial theorem?
To prove that $f(x)=x^p$ when $p=2$ what we do is the following.
Let $\epsilon=1$. Let $\delta >0$ be given. Now, let $x=\frac{1}{\delta}$ and $y=\frac{1}{\delta}+\frac{\delta}{2}$ in the common definition of uniformly continuous (see here). Then, $$|x-y|=\frac{\delta}{2}<\delta.$$
But $$|f(x)-f(y)|=|x-y||x+y|=|\frac{2}{\delta}+\frac{\delta}{2}|\frac{\delta}{2}> \frac{ 2\delta}{\delta 2}=1=\epsilon.$$
Thus, $f(x)=x^2$ is not uniformly continuous. Now, can you try with a general $p$?