<s>[INST] Limit of $\lim_{x \rightarrow 0}\frac{9}{x}\bigl(\frac{3}{(x+3)^3}-\frac{1}{9}\bigr)$ I have to determine the following:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)$$
I've got so far:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)= \lim_{x \rightarrow 0}\left(\frac{27}{x(x+3)^3}-\frac{1}{x}\right)=\lim_{x \rightarrow 0} \left(\frac{27-(x+3)^3}{x(x+3)^3}\right)=\cdots$$
How to go on? I've got $\frac{\infty}{0}...$
 [/INST] Just develop the nuumerator and you'll get 
$$
\lim_{x \rightarrow 0} \frac{-x^3 - 9x^2 - 27x}{x(x+3)^3}
=\lim_{x \rightarrow 0} \frac{-x^2 - 9x - 27}{(x +3)^3} = -1
$$
</s>