Câu 1: lim \(\frac{1^3+2^3+...+n^3}{n\left(n^3+1\right)}\)
Câu 2: lim (\(4+\frac{\left(-1\right)^n}{n+1}\) )
Câu 3: lim\(\sqrt{9-\frac{cos2n}{n}}\)
Câu 4: lim ( \(n^2sin\frac{n\pi}{5}-2n^3\))
Câu 5: Cho \(u_n=\frac{\left(-1\right)^n}{n^2+1}\) và \(v_n=\frac{1}{n^2+2}\). Khi đó tính lim \(\left(u_n+v_n\right)\)
Câu 4.
\(\lim \left( {{n^2}\sin \dfrac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - \infty \)
Vì \(\lim {n^3} = + \infty ;\lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2 \)
\(\left| {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n}} \right| \le \dfrac{1}{n};\lim \dfrac{1}{n} = 0 \Rightarrow \lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2\)
Câu 5.
Ta có: \(\left\{ \begin{array}{l} 0 \le \left| {{u_n}} \right| \le \dfrac{1}{{{n^2} + 1}} \le \dfrac{1}{n} \to 0\\ 0 \le \left| {{v_n}} \right| \le \dfrac{1}{{{n^2} + 2}} \le \dfrac{1}{n} \to 0 \end{array} \right. \to \lim {u_n} = \lim {v_n} = 0 \to \lim \left( {{u_n} + {v_n}} \right) = 0\)