<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] 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.
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