I am trying to prove: $$\lim_{x\to c} Ax^k = Ac^k.$$ What I have: for all $\epsilon > 0$ there exists $\delta > 0$ such that $0< |x−c|<\delta$ implies $|Ax^k−Ac^k|<\epsilon$. $$|Ax^k−Ac^k| = |A||x^k−c^k| < ???$$
I'm having trouble determining the connection between $|x^k−c^k|$ and $\delta$, any help would be appreciated.
I think it helps to stand back from the problem. Don’t try to get a formula for your desired delta, just look at the qualitative behavior.
What makes it all work is that your function $f(x)=Ax^k$ is surjective (every positive number is $A$ times a $k$-th power) and strictly monotone: increasing if $k>0$ and decreasing if $k<0$.
Remember that your task is, given an epsilon, to find a delta that does what you want. Well, let $Y=f(c)$, the value at the given $x$-value $c$. Consider $Y-\epsilon$ and $Y+\epsilon$. You want to guarantee that for $x$ close enough to $c$, $f(x)$ is in the open interval $\langle Y-\epsilon,Y+\epsilon\rangle$. Well, assuming $k>0$, so that your function is increasing, we’ll have $x^-<x<x^+$ with $f(x^-)=Y-\epsilon$ and $f(x^+)=Y+\epsilon$. Now just let $\delta$ be the smaller of $x-x^-$ and $x^+-x$. I’ll leave the rest to you.