Toán Giúp em bài này với ạ: Lim√4n^6+n+1/1-2n-3n^3 09/07/2021 By Lyla Giúp em bài này với ạ: Lim√4n^6+n+1/1-2n-3n^3
Đáp án: \[\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 – 2n – 3{n^3}}} = \frac{{ – 2}}{3}\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 – 2n – 3{n^3}}}\\ = \lim \frac{{\sqrt {{n^6}.\left( {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} \right)} }}{{{n^3}\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\ = \lim \frac{{{n^3}\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{{n^3}\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\ = \lim \frac{{\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\ = \frac{{\sqrt 4 }}{{ – 3}} = \frac{{ – 2}}{3}\end{array}\) Trả lời
Đáp án:
\[\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 – 2n – 3{n^3}}} = \frac{{ – 2}}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 – 2n – 3{n^3}}}\\
= \lim \frac{{\sqrt {{n^6}.\left( {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} \right)} }}{{{n^3}\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\
= \lim \frac{{{n^3}\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{{n^3}\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\
= \lim \frac{{\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{\left( {\frac{1}{{{n^3}}} – \frac{2}{{{n^2}}} – 3} \right)}}\\
= \frac{{\sqrt 4 }}{{ – 3}} = \frac{{ – 2}}{3}
\end{array}\)