To simplify the expression, we will first expand the terms inside the square roots:

√((n+2)(n+1)) - √((n+3)n)
= √(n^2 + 3n + 2) - √(n^2 + 3n)

Next, we can take out the common factors inside each square root:

= √(n^2 + 3n + 2) - √(n(n + 3))

Now, we can factorize the terms inside the square roots:

= √((n + 1)(n + 2)) - √(n(n + 3))

Since n is increasing without bound (n → ∞), the dominant term in each square root is the term with the highest power:

= √(n^2) - √(n^2)
= n - n
= 0

Therefore, lim(n → ∞) √((n+2)(n+1)) - √((n+3)n) = 0.