Tìm giới hạn sau: Lim ( (∛2n -n ³) + n -1 )

0 bình luận về “Tìm giới hạn sau: Lim ( (∛2n -n ³) + n -1 )”

1. Đáp án:

    

   Giải thích các bước giải:

   \(\begin{array}{l}
   \lim \sqrt[3]{{2n – {n^3}}} + n – 1 = \lim \frac{{\left( {\sqrt[3]{{2n – {n^3}}} + n – 1} \right)\left( {\sqrt[3]{{{{(2n – {n^3})}^2}}} – \sqrt[3]{{2n – {n^3}}}.\left( {n – 1} \right) + {{\left( {n – 1} \right)}^2}} \right)}}{{\sqrt[3]{{{{(2n – {n^3})}^2}}} – \sqrt[3]{{2n – {n^3}}}.\left( {n – 1} \right) + {{\left( {n – 1} \right)}^2}}}\\
    = \lim \frac{{2n – {n^3} + {n^3} – 3{n^2} + 3n – 1}}{{\sqrt[3]{{{{(2n – {n^3})}^2}}} – \sqrt[3]{{2n – {n^3}}}.\left( {n – 1} \right) + {{\left( {n – 1} \right)}^2}}}\\
    = \lim \frac{{ – 3{n^2} + 5n – 1}}{{\sqrt[3]{{{{(2n – {n^3})}^2}}} – \sqrt[3]{{2n – {n^3}}}.\left( {n – 1} \right) + {{\left( {n – 1} \right)}^2}}}\\
    = \lim \frac{{ – 3 + \frac{5}{n} – \frac{1}{{{n^2}}}}}{{\sqrt[3]{{\frac{4}{{{n^4}}} – \frac{4}{{{n^2}}} + 1}} – \sqrt[3]{{\frac{2}{{{n^2}}} – 1}}.\left( {1 – \frac{1}{n}} \right) + 1 – \frac{2}{n} + \frac{1}{{{n^2}}}}} = \frac{{ – 3}}{{1 + 1 + 1}} =  – 1
   \end{array}\)

   Bình luận