Đáp án:
B3:
c) \( - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)\lim \dfrac{{{{\left( {\dfrac{1}{3}} \right)}^n} + 1}}{{\dfrac{4}{{{3^n}}} + 1}} = 1\\
b)\lim \dfrac{{4.{{\left( {\dfrac{3}{7}} \right)}^n} + 7.1}}{{2.{{\left( {\dfrac{3}{7}} \right)}^n} + 1}} = 7\\
c)\lim \dfrac{{4.{{\left( {\dfrac{4}{8}} \right)}^n} + 36.{{\left( {\dfrac{6}{8}} \right)}^n}}}{{{{\left( {\dfrac{5}{8}} \right)}^n} + 1}} = 0\\
B3:\\
a)\lim \dfrac{{{n^2} + 3n - {n^2}}}{{\sqrt {{n^2} + 3n} + n}}\\
= \lim \dfrac{{3n}}{{\sqrt {{n^2} + 3n} + n}}\\
= \lim \dfrac{3}{{\sqrt {1 + \dfrac{3}{n}} + 1}} = \dfrac{3}{2}\\
b)\lim \dfrac{{{n^2} - 2n - {n^2} + 2.2013n - 2013}}{{\sqrt {{n^2} - 2n} + n - 2013}}\\
= \lim \dfrac{{4024n - 2013}}{{\sqrt {{n^2} - 2n} + n - 2013}}\\
= \lim \dfrac{{4024 - \dfrac{{2013}}{n}}}{{\sqrt {1 - \dfrac{2}{n}} + 1 - \dfrac{{2013}}{n}}} = \dfrac{{4024}}{2} = 2012\\
c)\lim \dfrac{{{n^2} - n - {n^2}}}{{\sqrt {{n^2} - n} + n}}\\
= \lim \dfrac{{ - n}}{{\sqrt {{n^2} - n} + n}}\\
= \lim \dfrac{{ - 1}}{{\sqrt {1 - \dfrac{1}{n}} + 1}} = - \dfrac{1}{2}
\end{array}\)
