I'm trying to prove that even for real exponent we have that $a^{x_1+x_2}=a^{x_1}a^{x_2}$, for every $x_1,x_2\in\mathbb{R}$ and $a>0$.
In other words, I have to show this:
$$\left(\lim_{\mathbb{Q}\ni r_1\to x_1}a^{r_1}\right) \cdot \left(\lim_{\mathbb{Q}\ni r_2\to x_2}a^{r_2}\right)=\left(\lim_{\mathbb{Q}\ni r\to x_1+x_2}a^{r}\right)$$
So, considering the left member, I can say that, given $\epsilon>0$, there exist $\delta_1$ and $\delta_2$ such that:
$$\left\{\begin{matrix} \left | r_1-x_1 \right |<\delta_1\Rightarrow \left | a^{x_1}-a^{r_1}\right |<\epsilon \\ \left | r_2-x_2 \right |<\delta_2\Rightarrow \left | a^{x_2}-a^{r_2}\right |<\epsilon \end{matrix}\right.$$
Multiplying the first inequality by $|a^{x_2}|$ and the second by $|a^{r_1}|$, I can assert that for any $\epsilon>0$ exists $\delta$ ($\delta=\min\{\delta_1,\delta_2\}$) such that:
$$\left | a^{x_1}a^{x_2}-a^{r_1+r_2} \right |<\epsilon\left ( |a^{r_1}|+|a^{x_2}| \right )$$
for any $r_1+r_2$ in a deleted neighborhood (in $\mathbb{Q}$) of $x_1+x_2$ with radius $\delta$.
I would terminate here the proof, but the quantity $\epsilon\left ( |a^{r_1}|+|a^{x_2}| \right )$ has a $r_1$ dependence. I can't remove it.
Does someone have an hint?
Thanks in advance.
Copy the demonstration of the limit of a product of functions : $$\left|a^x-a^{x_1}a^{x_2}\right| \le \left|a^x-a^{r_1+r_2}\right| + \left|a^{r_1+r_2}-a^{x_1}a^{x_2}\right|$$ As you claimed, you know how to get the first term lower than any $\epsilon>0$. Problem is the second term. \begin{align}\left|a^{r_1+r_2}-a^{x_1}a^{x_2}\right| &= \left|a^{r_1}a^{r_2}-a^{x_1}a^{x_2}\right| \\ &= \left|a^{r_1}a^{r_2}-a^{x_1}a^{r_2}+a^{x_1}a^{r_2}-a^{x_1}a^{x_2}\right| \\ &\le a^{r_2}\left|a^{r_1}-a^{x_1}\right| + a^{x_1}\left|a^{r_2}-a^{x_2}\right| \end{align} Now all you have to do is note $a^{x_1}$ is fixed, and as $a^{r_2}$ has limit $a^{x_2}$, it is bounded.
Is this what you are looking for ?