# From $\forall p \in \mathbb{R} \lim_{x \to 1} x^p = 1$ conclude that $\forall p \in \mathbb{R} \, f(x) := x^p$ is continuous on $(0, + \infty)$

I am trying to solve exercise 9.4.4 from Tao's Analysis I. It says:

Prove the following theorem:

Let $$p$$ be a real number. Then the function $$f : (0, \infty) \to \mathbb{R}$$ defined by $$f(x) := x^p$$ is continuous.

Hint: Prove that for all real numbers $$p$$ $$\lim_{x \to 1} x^p = 1$$. Then apply some basic facts about exponentiation.

The basic facts he refers to (he refers to the theorem by number) are:

Let $$x,y>0, q,r \in \mathbb{R}$$. Then:

1. $$x^q > 0$$
2. $$x^{q+r} = x^q x^r$$
3. $$(x^q)^r = x^{qr}$$
4. $$x^{-q} = 1 / x^q$$
5. if $$q>0$$, then $$x>y$$ iff $$x^q>y^q$$
6. If $$x>1$$, then $$x^q>x^r$$ iff $$q>r$$. If $$x<1$$, then $$x^q>x^r$$ iff $$q.

BTW the textbook defines exponentiation $$x^\alpha$$ for real $$\alpha$$ to be $$\lim_{n \to \infty} x^{q_n}$$, where $$(q_n)_{n=1}^\infty$$ is any sequence of rational numbers converging to $$\alpha$$.

I think I can prove it using other means - first proving that for any integer $$n$$ $$f(x) = x^n$$ is continuous, then prove that for any positive integer $$n$$ $$f(x) = x^{1/n}$$ is continuous, then prove that for any rational $$p$$ $$f(x) = x^p$$ is continuous, then prove continuity for real $$p$$.

But I have no idea how to do it using the author's hint. I have proved that $$\lim_{x \to 1} x^p = 1$$, but I've got no idea what to do with this fact. Please help me figure it out.

You want to show that, for every $$a>0$$, $$\lim_{x\to a} f(x) = f(a)$$ i.e., $$\lim_{x\to a} x^p = a^p$$ and you want to infer this from the case $$a=1$$. Observing that $${x^p}/{a^p} = (x/a)^p$$, can you prove that $$\lim_{x\to a} \frac{x^p}{a^p} = 1$$ and conclude from there?