Crossing the limits !

With the fact in mind that Product of Limits is the Limit of Product , we move on simplifying the expression:

$\begin{aligned} \left(\prod_{i=1}^5 A_i\right) \cdot \left(\prod_{j=1}^5 B_j\right) &=& \left(\prod_{i=1}^5 \lim_{x\rightarrow 0} \dfrac{(x+i)(x+i+1)}{x}\right) \cdot \left(\prod_{j=1}^5 \lim_{x\rightarrow 0} \dfrac{x}{x+j} \right) \\ \\ &=& \lim_{x\rightarrow 0} \left[\left(\prod_{i=1}^5 \dfrac{(x+i)(x+i+1)}{x}\right)\cdot \left(\prod_{j=1}^5 \dfrac{x}{(x+j)}\right)\right] \\ \\ &=& \lim_{x\rightarrow 0} \left[ \left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right) \right] \\ \\ &=& 2\times 3\times 4\times 5\times 6 \\ \\ &=& \fbox{720} \\ \end{aligned}$

Note that:

$\prod_{i=1}^5 A_i= \lim_{x\to 0} \frac{(x+1)(x+2)^2(x+3)^3(x+4)^2(x+5)^2(x+6)}{x^5}$

and

$\prod_{j=1}^5 B_j= \lim_{x\to 0} \frac{x^5}{(x+1)(x+2)(x+3)(x+4)(x+5)}$

Hence,

$(\prod_{i=1}^5 A_i ) \cdot (\prod_{j=1}^5 B_j)=\lim_{x\to 0} [\frac{(x+1)(x+2)^2(x+3)^3(x+4)^2(x+5)^2(x+6)}{x^5} \cdot \frac{x^5}{(x+1)(x+2)(x+3)(x+4)(x+5)} ]$

$\implies \lim_{x\to 0} [(x+2)(x+3)(x+4)(x+5)(x+6)]$

$\implies (0+2)(0+3)(0+4)(0+5)(0+6)$

$\implies 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6$

$\implies \boxed{720}.$