1. ## continuity question

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

using the $\displaystyle \delta \ \varepsilon$ yields:

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

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

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

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

using the $\displaystyle \delta \ \varepsilon$ yields:

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

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

$\displaystyle f(x)=x^n=(f \circ f \circ f \circ \dotso \circ f)(x)$. 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 $\displaystyle f$ is the $\displaystyle n$-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 $\displaystyle |x^n-c^n|=|x-c||x^{n-1}+x^{n-2}c +\cdots + x c^{n-2} + c^{n-1}|\leq |x-c| n |2c|^{n-1}$ for any [tex]x[/Math] such that [tex]|x|<2|c|[/Math]. I think you can conclude from there.

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

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