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Bạn xem lại câu a nhé! Làm gì phải là m2
b) \(lim\left(1+n^2-\sqrt{n^4+3n+1}\right)=lim\frac{\left(n^4+2n^2+1\right)-\left(n^4+3n+1\right)}{1+n^2+\sqrt{n^4+3n+1}}\)
\(=lim\frac{2n^2+3n}{1+n^2+\sqrt{n^4+3n+1}}=lim\frac{2+\frac{3}{n}}{\frac{1}{n^2}+1+\sqrt{1+\frac{3}{n}+\frac{1}{n^2}}}=\frac{2}{2}=1\)
c) = \(lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=0\)
d) = \(lim\frac{n+1}{\sqrt{n^2+n+1}+n}=lim\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}=\frac{1}{2}\)
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
a) \(\lim \frac{{ - 2n + 1}}{n} = \lim \frac{{n\left( { - 2 + \frac{1}{n}} \right)}}{n} = \lim \left( { - 2 + \frac{1}{n}} \right) = - 2\)
b) \(\lim \frac{{\sqrt {16{n^2} - 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {16 - \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {16 - \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {16 - \frac{2}{{{n^2}}}} = 4\)
c) \(\lim \frac{4}{{2n + 1}} = \lim \frac{4}{{n\left( {2 + \frac{1}{n}} \right)}} = \lim \left( {\frac{4}{n}.\frac{1}{{2 + \frac{1}{n}}}} \right) = \lim \frac{4}{n}.\lim \frac{1}{{2 + \frac{1}{n}}} = 0\)
d) \(\lim \frac{{{n^2} - 2n + 3}}{{2{n^2}}} = \lim \frac{{{n^2}\left( {1 - \frac{2}{n} + \frac{3}{{{n^2}}}} \right)}}{{2{n^2}}} = \lim \frac{{1 - \frac{2}{n} + \frac{3}{{{n^2}}}}}{2} = \frac{1}{2}\)
a;Chia n cả tử và mẫu
b;Chia cho n4 mà tử dần đến 0 mẫu dần đến 1 nên lim =0
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
Đây là ảnh của câu hỏi trên giải giúp với