- #1

Terrell

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## Homework Statement

Let ##p\in\Bbb{R}##. Then the function ##f:(0,\infty)\rightarrow \Bbb{R}## defined by ##f(x):=x^p##. Then ##f## is continuous.

I need someone to check what I've done so far and I really need help finishing the last part. I am clueless as to how to show continuity for arbitrary ##x_0\in\Bbb{R}##. I am following the hint given (see image), which asks me to apply proposition 6.7.3., but I don't see how that would complete the proof.

## Homework Equations

## The Attempt at a Solution

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__Show: ##\lim_{x\rightarrow 1}x^n=1##__*: ##n\in\Bbb{N}##; consider ##f(x)=x##.*

__CASE 1__**By 9.3.14**., ##\lim \limits_{x\to 1}(ff)(x)=\lim \limits_{x\to 1}f(x)\lim \limits_{x\to 1}f(x)=\lim \limits_{x\rightarrow 1}x\lim \limits_{x\to 1}x=1##.

*##n<0##; ##\lim \limits_{x\to 1}x^n=\lim \limits_{x\to 1}\frac{1}{x^{-n}}##. Since ##-n\in\Bbb{N}##, by 9.3.14., ##\lim \limits_{x\to 1}\frac{1}{x^{-n}}=\frac{\lim \limits_{x\to 1}1}{\lim \limits_{x\to 1}x^{-n}}=1##.*

__CASE 2__:

__Show: ##\lim \limits_{x\rightarrow 1}x^p=1## for all ##p\in\Bbb{R}##.__Note that ##\forall p\in\Bbb{R},\exists m,n\in\Bbb{Z}## such that ##m<p<n##. By proposition 6.7.3., if ##\vert x\vert<1## and ##m<p<n##, then ##x^n<x^p<x^m##. Thus, by the squeeze test ##\lim \limits_{x\to 1}x^m=1=\lim \limits_{x\to 1}x^n## implies ##\lim \limits_{x\to 1}x^p=1##. On the other hand, If ##\vert x\vert>1##, then, by proposition 6.7.3., ##m<p<n## implies ##x^m<x^p<x^n, \forall x\in\Bbb{R}##. Again by the squeeze test, ##\lim \limits_{x\to 1}x^m=1=\lim \limits_{x\to 1}x^n## implies ##\lim \limits_{x\to 1}x^p=1##.