<s>[INST] 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?
 [/INST] 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}
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