Hola
Me queda una expresión que a ver como se puede simplificar!
\( \displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{\displaystyle\frac{1}{3n+3}+\displaystyle\frac{1}{3n+2}+\displaystyle\frac{1}{3n+1}-\displaystyle\frac{1}{2n+1}-\displaystyle\frac{1}{2n+2}}{\displaystyle\frac{-1}{n(n-1)}}} \)
Pues hombre "remángate" y haz las cuentas que no es para tanto:
\( \displaystyle\frac{1}{3(n+1)}+\displaystyle\frac{1}{3n+2}+\displaystyle\frac{1}{3n+1}-\displaystyle\frac{1}{2n+1}-\displaystyle\frac{1}{2(n+1)}=
\displaystyle\frac{1}{3n+2}+\displaystyle\frac{1}{3n+1}-\displaystyle\frac{1}{2n+1}-\displaystyle\frac{1}{6(n+1)}=\\
=\dfrac{6(n+1)(3n+1)(2n+1)+6(n+1)(3n+2)(2n+1)-6(n+1)(3n+2)(3n+1)-(3n+2)(3n+1)(2n+1)}{6(3n+2)(3n+1)(2n+1)(n+1)}=\\
=\dfrac{9n^2+11n+4}{6(3n+2)(3n+1)(2n+1)(n+1)}\\ \)
Y ahora ya es inmediato terminar.
Saludos.