Proof Continuity of Power Functions

Question

I am trying to prove:

$$\lim_{x\to c}Ax^k=Ac^k.$$

What I have:

For $k > 0$

For all $\varepsilon >0$ there exists $\delta>0$ such that $0<|x−c|<\delta$ implies $|Ax^k−Ac^k|<\varepsilon$

$|Ax^k−Ac^k|=|A||x^k−c^k|<|A||(\delta-c)^k−c^k|=k.root((((|A||\varepsilon^k−c^k|)/|A|)+c^k))=\varepsilon$

I have doubts that I did this correctly.

Answer 1

If $k$ is an integer, you can use $x^k-c^k =(x-c)\sum_{j=0}^{k-1}x^jc^{k-1-j} $, let $x \to c$, and bound the sum.

Answer 2

you've noted $k>0$,

you could always Assume it is continuous and if you're wrong you should come to a contradiction.

then use tailors as a linear approximation. $$f(x)=\sum_{i=0}^{k}\frac{f^{i}(a)}{i!}(x-a)^{i}+\frac{r_k(x)}{k!}(x-a)^k$$ this gives $$Ax^{k}=Ac^{k}+Akc^{k-1}\frac{x-c}{1!}+...$$ $$Lim_{x\rightarrow c}Ax^k = lim_{x \rightarrow c} (Ac^{k}+Akc^{k-1}\frac{x-c}{1!}+...)$$ $$Lim_{x \rightarrow c}Ax^k=Ac^k$$ as req

Answer 3

A simpler approach would we show this directly:

$$\lim_{x\to c}Ax^k=Ac^k.$$

Notice that $\lim_{x\to c}Ax^k$ is just $(\lim_{x\to c}x) * (\lim_{x\to c}x)*(\lim_{x\to c}x) \dots \text{ k times, and we can do this because it's the product rule = $x^k$}$

So we can say that: $\lim_{x\to c}Ax^k$ is just $(\lim_{x\to c}x) * (\lim_{x\to c}x)*(\lim_{x\to c}x) \dots = c*c*c \dots \text{ this goes k times = $c^k$}$

Then we know from constant limit, that we can just add our constant $A$.

Using epsilon-delta is kinda hard for this example.