1. ## continuity question

Show that $f(x)=x^n$ where $n \in \mathbb{N}$ is continuous on the interval $[a,b]$.

using the $\delta \ \varepsilon$ yields:

$\forall \ \delta>0, \ \varepsilon >0 \ \ |x-c|<\delta, \ \mbox{then} \ |x^n-c^n|<\varepsilon$ for $n \in \mathbb{N}$ at which point I don't know how to continue.

I figure, I can write $f(x)= x\cdot x \cdot x \dotso \cdot x$. furthermore I can express this as a number of composite functions where I would have:

$f(x)=x^n=(f \circ f \circ f \circ \dotso \circ f)(x)$.

I know how to show that given two continuous function then their composite is continuous, but I don't know how I would show that n composite functions are continuous, I imagine this would have to be done by induction.

2. Originally Posted by lllll
Show that $f(x)=x^n$ where $n \in \mathbb{N}$ is continuous on the interval $[a,b]$.

using the $\delta \ \varepsilon$ yields:

$\forall \ \delta>0, \ \varepsilon >0 \ \ |x-c|<\delta, \ \mbox{then} \ |x^n-c^n|<\varepsilon$ for $n \in \mathbb{N}$ at which point I don't know how to continue.

I figure, I can write $f(x)= x\cdot x \cdot x \dotso \cdot x$. furthermore I can express this as a number of composite functions where I would have:

$f(x)=x^n=(f \circ f \circ f \circ \dotso \circ f)(x)$. This is wrong!

I know how to show that given two continuous function then their composite is continuous, but I don't know how I would show that n composite functions are continuous, I imagine this would have to be done by induction.
There's no reason (of possibility) for introducing composite functions here. The function $f$ is the $n$-th power of a continuous function, and hence is continuous. But this is exactly what you need to prove, so here is a hint for the proof:

The key inequality is $|x^n-c^n|=|x-c||x^{n-1}+x^{n-2}c +\cdots + x c^{n-2} + c^{n-1}|\leq |x-c| n |2c|^{n-1}$ for any $$x$$ such that $$|x|<2|c|$$. I think you can conclude from there.

3. The only part I'm a little confused about, is your statement that $|x| <2|c|$.

4. Originally Posted by lllll
The only part I'm a little confused about, is your statement that $|x| <2|c|$.
I wrote this for brevity, but in fact, I only need $x$ not to be too large, in order to bound $|x|$, but not too close to 0 since $x$ is supposed to be about equal to $c$. For instance, I could have said "Choose any $M>|c|$. If $|x|\leq M$, then $|x^k c^{n-1-k}|\leq M^k |c|^{n-1-k}\leq A^{n-1}$ where $A=\max(M,|c|,1)$, so that $|x^n-c^n|=|x-c||x^{n-1}+x^{n-2}c +\cdots + x c^{n-2} + c^{n-1}|\leq |x-c|n A^{n-1}$. " Then, when you choose $\delta$, it must be such that $\delta n A^{n-1}\leq \varepsilon$ and $\delta\leq M-|c|$, so that $|x-c|<\delta$ implies $|x|< |c|+\delta\leq M$ and we can apply the above.