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?
 A: 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}
A: Try multiplying out the top and factoring out x from the top and bottom...
Hint: Top = $(x^4+10 x^3+35 x^2+50 x)$
A: L'Hopital's rule is not that painful. For the numerator, you have to differentiate a product of linear terms. No need to expand out; instead, use the product rule for a product of four factors:
$$
(uvwz)'=u'vwz + uv'wz + uvw'z + uvwz'
$$
In your case the derivative of $x+c$ is one, so the derivative of the numerator amounts to the sum of entities that look like the original product but with one term deleted:
$$
(x+2)(x+3)(x+4) + (x+1)(x+3)(x+4) + (x+1)(x+2)(x+4) + (x+1)(x+2)(x+3)
$$
Plugging in $x=0$ gets you the familiar value $2\cdot3\cdot4 + 1\cdot3\cdot4 + 1\cdot2\cdot4+1\cdot2\cdot3=50$.
A: 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.
A: $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}
$
More generally,
if
$F(x)
=\prod_{k=1}^m f_k(x)
$,
then
$\ln F(x)
=\sum_{k=1}^m \ln f_k(x)
$
so that
$\dfrac{F'(x)}{F(x)}
=(\ln F(x))
=\sum_{k=1}^m (\ln f_k(x))'
=\sum_{k=1}^m \dfrac{f_k'(x)}{f_k(x)}
$.
Therefore
$F'(x)
=F(x)\sum_{k=1}^m \dfrac{f_k'(x)}{f_k(x)}
=\sum_{k=1}^m f_k'(x)\prod_{j=1, j\ne k}^m f_j(x)
$.
If the $f_k$ are all linear functions
of the form
$f_k(x) = x+a_k$,
then
$f_k'(x) = 1$
and
$f_k(0) = a_k$,
so
$F'(x)
=\sum_{k=1}^m F(x)\dfrac{1}{f_k(x)}
=\sum_{k=1}^m \prod_{j=1, j\ne k}^m f_j(x)
$.
Therefore
$\begin{array}\\
F'(0)
&=F(0)\sum_{k=1}^m \dfrac{1}{f_k(0)}\\
&=\prod_{k=1}^m a_k\sum_{k=1}^m \dfrac{1}{a_k}\\
&=\sum_{k=1}^m \prod_{j=1, j\ne k}^m f_j(0)
\qquad\text{(another representation)}\\
&=\sum_{k=1}^m \prod_{j=1, j\ne k}^m a_j\\
\end{array}
$
