Definition: The base $a$ logarithm ($a\in]0,1[\cup]1,+\infty[$) is the continuous function defined by: $\log_a(xy)=\log_a(x)+\log_a(y)~~\forall x,y>0$ and $\log_a(a)=1$
If I used this definition of the logarithm how can deduce the derivative of $\log_a(x)$ and $\lim\limits_{x\to +\infty} \log_a(x)$? Is the continuity necessary in the definition or should I replace it with another property (weaker or stronger) like uniform continuity, differentiability, Holder-continuity, etc.
On
For $x\ne0$, we have (with $h'=hx$): $$\lim_{h'\to0}\frac{\log_a(x+h')-\log_a(x)}{h'}=\lim_{h\to0}\frac{\log_a(x+hx)-\log_a(x)}{hx}=\frac1x\lim_{h\to0}\frac{\log_a(x(1+h))-\log_a(x)}h.$$
And by the stated definition, $$\lim_{h\to0}\frac{\log_a(x(1+h))-\log_a(x)}h=\lim_{h\to0}\frac{\log_a x+\log_a(1+h)-\log_a(x)}h=\lim_{h\to0}\frac{\log_a(1+h)}h,$$ a function of $a$ alone, so that $$(\log_a x)'=\frac{C(a)}x.$$
If the last limit exists, then the function is differentiable for any $x$.
Here I have a partial solution for derivative of Logarithms
(not only using the definition, I have assume the continuity).
By the Well known limit $$\lim_{n\to\infty}\Big(1+\dfrac{x}{n}\Big)^n=e^x,$$ we can drive $$\lim_{\delta x\to 0}\Big(1+\dfrac{\delta x}{x}\Big)^{\dfrac{1}{\delta x}}=e^{\dfrac{1}{x}}.$$ Since Logarithm is a continuous function, taking the logarithm of both sides, we can obtain $$\lim_{\delta x\to 0}\dfrac{\log_a\Big(1+\dfrac{\delta x}{x}\Big)}{\Big(\dfrac{\delta x}{x}\Big)}=\log_ae$$
Let $f(x)=\log_ax,$ then $$\dfrac{f(x+\delta x)-f(x)}{\delta x}=\dfrac{\log_a(x+\delta x)-\log_ax}{\delta x}=\dfrac{\log_a\Big(1+\dfrac{\delta x}{x}\Big)}{\Big(\dfrac{\delta x}{x}\Big)}\dfrac{1}{x}$$ $$f'(x)=\lim_{\delta x\to0}\dfrac{f(x+\delta x)-f(x)}{\delta x}=\lim_{\delta x\to0}\dfrac{\log_a\Big(1+\dfrac{\delta x}{x}\Big)}{\Big(\dfrac{\delta x}{x}\Big)}\lim_{\delta x\to0}\dfrac{1}{x}$$ $$f'(x)=\dfrac{\log_ae}{x}$$