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 $\frac{\infty}{0}...$

When you plug in $0$ to $x$, you see that the answer is $0/0$. You have to use L'Hospital's Rule, which says
$$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}.$$
This applies only to $0/0$ or $\infty/\infty$.
Hence, you just need to take the derivative of the top and bottom until you get an answer that is either the answer or you can't use L'Hospital's anymore.
So let's take the derivatives of the numerator and denominator:
$$\frac{d}{dx}\left(27-(x+3)^3\right)= -3(x+3)^2$$
$$\frac{d}{dx}\left(x(x+3)^3\right) = x(3(x+3)^2) + (x+3)^3$$
So now we can just plug zero in!
$$\lim_{x\to 0} \frac{-3(x+3)^2}{3x(x+3)^2 + (x+3)^3} = -27/27 = -1$$
