Hi,
I do not know how to solve below problem; maybe someone can guide me through it:
Prove that $\displaystyle f(x)=x^n$ is continuous on $\displaystyle R$ for any positive integer $\displaystyle n$.
Thank you.
It's easy to show that $\displaystyle f(x)=x^n$ is not only continuous, but uniformly continuous on every closed segment $\displaystyle [-C,C]$. Indeed, fix a $\displaystyle C$ and assume $\displaystyle x_1,x_2\in[-C,C]$. We have
$\displaystyle x_1^n-x_2^n =(x_1-x_2) (x_1^{n-1}+ x_1^{n-2}x_2+ \dots+x_1x_2^{n-2}+ x_2^{n-1})$
(this is checked by multiplying the right-hand side). The second factor in the right-hand side has $\displaystyle n$ terms and $\displaystyle |x_1^{n-i}x_2^{i-1}|\le C^{n-1}$. Therefore, $\displaystyle |x_1^n-x_2^n|\le nC^{n-1}|x_1-x_2|$. The expression $\displaystyle nC^{n-1}$ is a constant for each fixed $\displaystyle n$, so a small variation in $\displaystyle x$ produces only a small variation in $\displaystyle f(x)$. It's easy to turn this into an $\displaystyle \varepsilon-\delta$ proof.