Kết quả của giới hạn lim ( 1 1 .4 + 1 2 .5 + . . . + 1 n ( n + 3 ) ) là:

\[\lim \left( {\frac{{\rm{1}}}{{{\rm{1}}{\rm{.4}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ + }}...{\rm{ + }}\frac{1}{{{\rm{n}}\left( {{\rm{n + 3}}} \right)}}} \right)\]

\[ = \lim \frac{1}{3}\left( {1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{6} + ... + \frac{1}{{\rm{n}}} - \frac{1}{{{\rm{n}} + 3}}} \right)\]

\[ = \lim \frac{1}{3}\left[ {\left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{{\rm{1}}}{{\rm{n}}}} \right) - \left( {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... + \frac{1}{{{\rm{n}} + 3}}} \right)} \right]\]

\[ = \lim \frac{1}{3}\left( {1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{{{\rm{n}} + 1}} - \frac{1}{{{\rm{n}} + 2}} - \frac{1}{{{\rm{n}} + 3}}} \right)\]

\[ = \lim \frac{1}{3}\left( {\frac{{11}}{6} - \frac{1}{{{\rm{n}} + 1}} - \frac{1}{{{\rm{n}} + 2}} - \frac{1}{{{\rm{n}} + 3}}} \right) = \frac{{11}}{{18}}\]

Đáp án cần chọn là: A