Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$ Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$
For this I think I should use De l'Hopital's rule but it takes a lot time and I can't get to answer.
Can we use the De l'Hopital's rule twice?or three times?If yes what is the limit of that?and how can we find the limit of polynomial multiplys?

Since we have a limit at $0$ of polynomials, only the term with the least degree matters. We have 
$$
(x+1)(x+2)(x+3)(x+4)-24=24+(2\times3\times4+1\times3\times4+1\times2\times3+1\times2\times4)x+c_2x^2+c_3x^3+c_4x^4-24=50x+c_2x^2+c_3x^3+c_4x^4.
$$
Then
\begin{align}
\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}
&=\frac{50x+c_2x^2+c_3x^3+c_4x^4}{x(x+5)}\\ \ \\
&=\frac{50}{x+5}+\frac{c_2x}{x+5}+\frac{c_3x^2}{x+5}+\frac{c_4x^3}{x+5}\\ \ \\
&\to\frac{50}5=10.
\end{align}
