Continuity of $f(x)=x^p$ when $p$ is a real number and $x\in (0,\infty)$ Here is my final answer. 
Definition Let $x>0$ be a real, and $\alpha$ be a real number. We define the quantity $x^{\alpha}$, by the formula $\text{lim}_{n\rightarrow\infty} x^{q_n}$ where $(q_n)$ is a sequence of rationals which converges to $\alpha$. (I've shown that the definition is well-defined).
Proposition: Let $p$ be a real number. Then the function $f: (0,\infty)\rightarrow \mathbb{R}$ defined by $f(x)=x^p$ is continuous. 
Proof: Let $x_0\in (0,\infty)$ we have to show  $\lim_{x\rightarrow x_0, x\in(0,\infty)} f(x)=f(x_0)$.
(1) Suppose $x_0=1$. 

Claim 1: For all natural numbers $\lim_{x\rightarrow 1, x\in\mathbb{R}} x^n=1$. Let $n=0$, so $x^n=1$ which is trivial. Suppose we have  proven the assertion for $n\ge 0$. So, $x^{n+1}=x^nx$ and then 
$$\lim_{x\rightarrow 1, x\in\mathbb{R}} x^{n}x=\lim_{x\rightarrow 1, x\in\mathbb{R}} x^{n}\cdot\lim_{x\rightarrow 1, x\in\mathbb{R}} x=1$$

2) Now we have to show that $\lim _{x\rightarrow 1; x\in (0, \infty)} x^p = 1$.
Let  $(x_n)_{n=0}^\infty$ be a sequence of positive real numbers which converges to $1$. We' like to show that $(x_n^p)\rightarrow 1$. 
Let $\varepsilon>0$ be arbitrary and let choose some $m\in \mathbb{N}$ so that $m> p$. Since $(1+1/k)_{k=1}^\infty$ and $(1-1/k)_{k=1}^\infty$ both converges to $1$, using the claim $1$, $(1+1/k)^m$ and $(1-1/k)^m$ converge also to $1$. Let $K_\varepsilon$ be a natural number such that both sequences are $\varepsilon$- close to $1$ for any $k\ge K_\varepsilon$. Let us fix some $k$, so that  $k\ge K_\varepsilon$ and $1-1/k > 0$.
Since $(x_n)$ converges to $1$, there is some $N_{1/k}$ such that $|x_n-1|\le 1/k$ for all $n\ge N_{1/k}$, i.e., $1-1/k\le x_n\le 1+1/k$. So, $(1-1/k)^p\le x_n^p\le (1+1/k)^p$. Also we know that $1+1/k>1$ and $p<m$, thus $(1+1/k)^p<(1+1/k)^m$. Similarly $1-1/k<1$, thus  $(1-1/k)^p>(1-1/k)^m$. Putting all the inequalities together we have 
$$(1-1/k)^m< x_n^p< (1+1/k)^m $$
Since both $(1+1/k)^m $ and $(1-1/k)^m $ are $\varepsilon$-close to $1$, hence $x_n^p$ is. Thus $(x_n)^p$ converges to $1$ as desired. Now since $(x_n)$ was an arbitrary sequence of positive real numbers converging to $1$, hence the result  hold for any sequences. Therefore  $\lim _{x\rightarrow 1; x\in (0, \infty)} x^p = 1$.
3) Let $x_0 \in (0 \infty)\backslash\{1\}$ and let $(x_n)_{n=0}^\infty$ be a sequence of positive real numbers which converges to $x_0$. Using the limit laws we know that $x_n/x_0$ converges to $1$ and so by part 2), we have $(x_n/x_0)^p \rightarrow 1$. Thus
\begin{align}\lim_{n\rightarrow \infty}x_n^p=\lim_{n\rightarrow \infty} x_0^p (x_n/x_0)^p\\
= x_0^p \lim_{n\rightarrow \infty} (x_n/x_0)^p \end{align}
and since $\lim_{n\rightarrow \infty} (x_n/x_0)^p=1 $, hence $\lim_{n\rightarrow \infty}x_n^p=x_0^p$. Since $(x_n)$ was an arbitrary sequence of positive real numbers converging to $x_n$, this would imply that $f$ is continuous on $(0,\infty)$ as desired.
Thanks to anyone. 
 A: The book's hint is spot on: suppose you have proven that $x^p$ is continuous at $1$; and choose some $\alpha>0$ distinct from $1$. Prove that $x^p$ is continuous at $\alpha$ iff $\alpha^{-p}x^p$ is continuous at $\alpha$, and note that this last is equivalent to $$\left(\frac x\alpha\right)^p\to 1$$ as $x\to \alpha$. But $x/\alpha\to 1$ as $x\to \alpha$, and since $t^p$ is continuous at $t=1$; we're done! You do have to prove $(xy)^p=x^py^p$ for $x,y$ real numbers first. 
A: Sorry for put it here as answer but I think is not necessary to open other question because is related, hopefully this is correct if not, I apologize. 
I'm assume is correct what with the help of Pedro do, it'd be great is someone could say me if is valid the reasoning behind (1). 
On the other hand I'd like to show $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by the formula $f(x)= a^x$ where $a\in \mathbb{R}^{>0}$ is continuous on $\mathbb{R}$. 
I think this is better :) (using a similar trick)
Suppose $a>1$ and $x=0$.
Let $(x_n)$ be a sequence real numbers such that $x_n\rightarrow 0$. For sake of contradiction suppose $(a^{x_n})\not\rightarrow 1$. Then, there is an $\varepsilon>0$ such that for all $\delta>0$ exists a $x_n$ so that  $|x_n|<\delta$ and $|a^{x_n}-1|>\varepsilon$. We set
$$B_j=\{j>j-1: |x_j|\le 1/j \text{ and }\, |a^{x_j}-1|>\varepsilon\,\}$$
Clearly this set is non-empty for all $j\ge 1$ since otherwise contradicts our assumption. Define $b_j = \min (B_j)$. Then $(b_j)\rightarrow 0$, also for all $j$ we have $-1/j\le b_j \le1/j$. Since $a>1$, we have $a^{-1/j}\le a^{b_j}\le a^{1/j}$. We already know that $(a^{1/j})$ and $(a^{-1/j})$ converges to $1$, so by the squeeze theorem we have $a^{x_j}\rightarrow 1$. But since $x_j\in X_j$, we have that $|a^{x_j}-1|>\varepsilon$. Contradiction.
If $a=1$, the result is trivial. If $a<1$, then $\lim_{x\rightarrow 0; x\in \mathbb{R}}a^x=\lim_{x\rightarrow 0; x\in \mathbb{R}}1/(1/a)^x$. Since $1/a>1$ the result follows by the limit laws and for what we have shown.
Now suppose $x_0\not = 0$. We'd like to show that $\lim_ {x\rightarrow x_0}a^x =a ^{x_0}$. Let $x\rightarrow x_0$, so $x-x_0\rightarrow 0$ and for what we have proven  $a^{x-x_0}\rightarrow 1$. Since $\,a^{x-x_0}=a^{x}a^{-x_0}$, we're done.
