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?

The limit above gives: $$\lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}=\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}$$
Now, it's enough to do the classical polynomial division, so we have: $$\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}=\lim\limits_{x\to 0}x^2+5x+10=10\hspace{75pt} (*1)$$
You can also factorize $x^4+10x^3+35x^2+50x$, whose two solutions are $x_1=0$ and $x_2=-5$, so we have $x^4+10x^3+35x^2+50x=x(x+5)(x^2+5x+10)$. By simplifying, we got the limit in $(*1)$.
I hope I told you what you were looking for.
