Proof that the exponential function is continuous on $\mathbb{R}$ without use of derivatives I am still trying to understand how to prove statements. I want to prove that for $a>0$, $f(x) = a^x$ is continuous on $\mathbb{R}$. The text gives an hint, namely, that it suffices proving continuity at 0.
Now, using the definition of continuity of a function, am I wrong if I write:
IF $|a^x - 0| < \epsilon$,  THEN $|x-0| < \delta$
If I have got something, I have to find an equation which relates epsilon and delta, which satisfies the definition. Actually, I ask this question because I am still not sure if I have understood how to prove using the formal definition of limit and continuity of a function.
PS. I want to prove the statement without making use of derivatives or even more complex arguments.
 A: Your definition is wrong.
Note that , for $a >0$, we have $a^0=1$ , so to prove the continuity of the function at $x=0$ you have to show that:

For all $\epsilon >0$ there is a $\delta >0$ such that  $$
 |x-0|<\delta \Rightarrow |a^x-1|<\epsilon$$

Can you do this?

The  point is to show that the solution of the inequality 
$(1) \qquad|a^x-1|<\epsilon$
is $|x|<\delta$ for some $\delta >0$.
So we can start solving $(1)$ and this require a bit of attention. The first step is to eliminate the absolute value, so we have to find where $a^x-1>0 \iff a^x>1$. The solution of this inequality is different for $a>1$ or $a<1$ ( the case $a=1$ is trivial). For $a>1$ we find the solution $x>0$, for $a<1$ we find $x<0$.
Consider the case $a>1$. The inequality $(1)$ becomes the two systems
$$
\begin{cases}
x>0\\
a^x<1+\epsilon
\end{cases}
\quad \lor \quad
\begin{cases}
x<0\\
a^x>1-\epsilon
\end{cases}
$$
Now we can solve these systems (using logarithms), and find that the solution is a neighborhood of $0$.
For the first system we have:
$$
a^x<1+\epsilon \iff e^{x\ln a}<e^{\ln(1+\epsilon)} \iff x\ln a<\ln(1+\epsilon) \iff x<\frac{\ln(1+\epsilon)}{\ln a}
$$
( note that $\ln a>0$ since $a>1$), so the solution is:
$$
0<x<\frac{\ln(1+\epsilon)}{\ln a}
$$
In a similar way we solve the second system and we find
$$
\frac{\ln(1-\epsilon)}{\ln a}<x<0
$$
so the solution of $(1)$ for $a>1$ is
$$
\frac{\ln(1-\epsilon)}{\ln a}<x<\frac{\ln(1+\epsilon)}{\ln a}
$$
and we can chose 
$$\delta=\mbox{Min}\left(\left|\frac{\ln(1+\epsilon)}{\ln a}\right|,\left|\frac{\ln(1-\epsilon)}{\ln a}\right|\right)$$
Than we have to do the same for the case $a<1$.
