Since $f^{\text{'}}(x)=-\frac{9}{1-9x}$ and $g^{\text{'}}(x)=1,$ applying L'Hospital's Rule and finding the resulting limit gives us the following.

$\underset{xa}{lim}\frac{f(x)}{g(x)}=\underset{xa}{lim}\frac{t^{\text{'}}(x)}{g^{\text{'}}(x)}$

$\underset{x0}{lim}\frac{ln(1-9x)}{x}=\underset{x0}{lim}\frac{\text{E}\text{n}\text{t}\text{e}\text{r}an\text{e}\text{x}\text{a}\text{c}\text{t}\text{n}\text{u}\text{m}\text{b}\text{e}\text{r}\text{.}}{1}$

$=\underset{x0}{lim}(-\frac{1-9x}{1})$

$=$