- LG a
- LG b
- LG c
- LG d
Tính các giới hạn sau:
LG a
\[\lim \sqrt {3{n^4} - 10n + 12} \]
Lời giải chi tiết:
\[\lim \sqrt {3{n^4} - 10n + 12} \] \[= \lim {n^2}.\sqrt {3 - {{10} \over {{n^3}}} + {{12} \over {{n^4}}}} \] \[= + \infty \]
Vì
\[\left\{ \begin{array}{l}
\lim {n^2} = + \infty \\
\lim \sqrt {3 - \frac{{10}}{{{n^3}}} + \frac{{12}}{{{n^4}}}} = \sqrt 3 > 0
\end{array} \right.\]
LG b
\[\lim \left[ {{{2.3}^n} - {{5.4}^n}} \right]\]
Lời giải chi tiết:
\[\lim \left[ {{{2.3}^n} - {{5.4}^n}} \right] \] \[= \lim {4^n}\left[ {2{{\left[ {{3 \over 4}} \right]}^n} - 5} \right] = - \infty \]
Vì
\[\left\{ \begin{array}{l}
\lim {4^n} = + \infty \\
\lim \left[ {2.{{\left[ {\frac{3}{4}} \right]}^n} - 5} \right] = - 5 < 0
\end{array} \right.\]
LG c
\[\lim \left[ {\sqrt {{n^4} + {n^2} + 1} - {n^2}} \right]\]
Lời giải chi tiết:
\[\eqalign{ & \lim \left[ {\sqrt {{n^4} + {n^2} + 1} - {n^2}} \right] \cr&= \lim \frac{{\left[ {\sqrt {{n^4} + {n^2} + 1} - {n^2}} \right]\left[ {\sqrt {{n^4} + {n^2} + 1} + {n^2}} \right]}}{{\sqrt {{n^4} + {n^2} + 1} + {n^2}}} \cr &= \lim \frac{{{n^4} + {n^2} + 1 - {n^4}}}{{\sqrt {{n^4} + {n^2} + 1} + {n^2}}}\cr &= \lim {{{n^2} + 1} \over {\sqrt {{n^4} + {n^2} + 1} + {n^2}}} \cr &= \lim \frac{{{n^2} + 1}}{{\sqrt {{n^4}\left[ {1 + \frac{1}{{{n^2}}} + \frac{1}{{{n^4}}}} \right]} + {n^2}}} \cr & = \lim \frac{{{n^2}\left[ {1 + \frac{1}{{{n^2}}}} \right]}}{{{n^2}\sqrt {1 + \frac{1}{{{n^2}}} + \frac{1}{{{n^4}}}} + {n^2}}} \cr & = \lim \frac{{{n^2}\left[ {1 + \frac{1}{{{n^2}}}} \right]}}{{{n^2}\left[ {\sqrt {1 + \frac{1}{{{n^2}}} + \frac{1}{{{n^4}}}} + 1} \right]}}\cr &= \lim {{1 + {1 \over {{n^2}}}} \over {\sqrt {1 + {1 \over {{n^2}}} + {1 \over {{n^4}}}} + 1}} \cr &= \frac{{1 + 0}}{{\sqrt {1 + 0 + 0} + 1}}= {1 \over 2} \cr} \]
LG d
\[\lim {1 \over {\sqrt {{n^2} + 2n} - n}}\]
Lời giải chi tiết: