Câu 3
\(lim\dfrac{2n^2+1}{3n^3-3n+3}=lim\dfrac{\dfrac{2n^2}{n^3}+\dfrac{1}{n^3}}{3-\dfrac{3n}{n^3}+\dfrac{3}{n^3}}=0\)
\(lim\dfrac{n\sqrt{n^2+1}}{\sqrt{4n^4-n^2+3}}=lim\dfrac{n\left|n\right|\sqrt{1+\dfrac{1}{n}}}{\left|n^2\right|\sqrt{4-\dfrac{2}{n^2}+\dfrac{3}{n^4}}}=1\)
Với n -> + vô cùng => lim bth = 1
Với n -> - vô cùng => lim bth = -1
a, sai
b, sai
c, \(cosx=0\Leftrightarrow cosx=cos\left(\dfrac{\pi}{2}\right)\Rightarrow x=\dfrac{\pi}{2}\)đúng
d, TH1 \(u_3=u_1+2d\Rightarrow\dfrac{3}{2}=u_1+2\Rightarrow u_1=\dfrac{3}{2}-2=-\dfrac{1}{2}\)( vô lí )
TH2 : \(\dfrac{3}{2}=u_1-2\Rightarrow u_1=\dfrac{3}{2}+2=\dfrac{7}{2}\)( vô lí )
=> sai
bài này là đúng/ sai hả bn ?
\(lim\dfrac{5n^3-2n+1}{n-2n^3}=lim\dfrac{\dfrac{5n^3}{n^3}-\dfrac{2n}{n^3}+\dfrac{1}{n^3}}{\dfrac{n}{n^3}-\dfrac{2n^3}{n^3}}=lim\dfrac{5-\dfrac{2}{n^2}+\dfrac{1}{n^3}}{\dfrac{1}{n^2}-2}=-\dfrac{5}{2}\)
a, đúng
b, sai
c, đúng do -1 =< sinx =< 1
d, \(u_3=u_1+2d=-\dfrac{5}{2}+2d\Rightarrow6=-\dfrac{5}{2}+2d\Rightarrow d=\dfrac{17}{4}\)sai
