Giải bài 4.4 trang 156 - SBT Đại số và Giải tích lớp 11
Tính giới hạn của các dãy số có số hạng tổng quát sau đây, khi \(n\to \infty\).
\(a)\,{{a}_{n}}=\dfrac{2n-3{{n}^{3}}+1}{{{n}^{3}}+{{n}^{2}}}\)
\(b)\,{{b}_{n}}=\dfrac{3{{n}^{3}}-5n+1}{{{n}^{2}}+4}\)
\(c)\,{{c}_{n}}=\dfrac{2n\sqrt{n}}{{{n}^{2}}+2n-1}\)
\(d)\,{{u}_{n}}={{2}^{n}}+\dfrac{1}{n}\)
\(e)\,{{v}_{n}}={{\left( -\dfrac{\sqrt{2}}{\pi } \right)}^{n}}+\dfrac{{{3}^{n}}}{{{4}^{n}}}\)
\(f)\,{{u}_{n}}=\dfrac{{{3}^{n}}-{{4}^{n}}+1}{{{2.4}^{n}}+{{2}^{n}}} \)
\(g)\,{{v}_{n}}=\dfrac{\sqrt{{{n}^{2}}+n-1}-\sqrt{4{{n}^{2}}-2}}{n+3}\)
a)
\(\lim \dfrac{2n-3{{n}^{3}}+1}{{{n}^{3}}+{{n}^{2}}}=\lim \dfrac{-3+\dfrac{2}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}}}{1+\dfrac{1}{n}}=-3 \)
Vì \(\lim \left( -3+\dfrac{2}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}} \right)=-3;\lim \left( 1+\dfrac{1}{n} \right)=1 \)
b) \( \lim \dfrac{3{{n}^{3}}-5n+1}{{{n}^{2}}+4}=\lim \dfrac{3-\dfrac{5}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}}}{\dfrac{1}{n}+\dfrac{4}{{{n}^{3}}}}=+\infty \)
Vì \(\lim \left( 3-\dfrac{5}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}} \right)=3>0\) và \(\lim \left( \dfrac{1}{n}+\dfrac{4}{{{n}^{3}}} \right)=0 \)
c) \( \lim \dfrac{2n\sqrt{n}}{{{n}^{2}}+2n-1}=\lim \dfrac{\dfrac{\sqrt{2}}{\sqrt{n}}}{1+\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}}}=0 \)
Vì \(\lim \dfrac{2}{\sqrt{n}}=0,\lim \left( 1+\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}} \right)=1>0 \)
d) \(\lim \left( {{2}^{n}}+\dfrac{1}{n} \right)=\lim {{2}^{n}}+\lim \dfrac{1}{n}=+\infty \)
e) \(\lim \left[ {{\left( -\dfrac{\sqrt{2}}{\pi } \right)}^{n}}+\dfrac{{{3}^{n}}}{{{4}^{n}}} \right]=\lim {{\left( -\dfrac{\sqrt{2}}{\pi } \right)}^{n}}+\lim {{\left( \dfrac{3}{4} \right)}^{n}}=0\)
Vì \(\left| -\dfrac{\sqrt{2}}{\pi } \right|<1\Rightarrow \lim \left( -\dfrac{\sqrt{2}}{\pi } \right)^n=0 \)
Và \(\dfrac{3}{4}<1\Rightarrow \lim {{\left( \dfrac{3}{4} \right)}^{n}}=0 \)
f) \(\lim \dfrac{{{3}^{n}}-{{4}^{n}}+1}{{{2.4}^{n}}+{{2}^{n}}}=\lim \dfrac{{{\left( \dfrac{3}{4} \right)}^{n}}+{{\left( \dfrac{1}{4} \right)}^{n}}-1}{2+{{\left( \dfrac{2}{4} \right)}^{n}}}=-\dfrac{1}{2} \)
g)
\( \begin{align} & \lim \dfrac{\sqrt{{{n}^{2}}+n-1}-\sqrt{4{{n}^{2}}-2}}{n+3} \\ & =\lim \dfrac{n\sqrt{1+\dfrac{1}{n}-\dfrac{1}{{{n}^{2}}}}-2n\sqrt{1-\dfrac{2}{4{{n}^{2}}}}}{n\left( 1+\dfrac{3}{n} \right)}\\&=\lim \dfrac{\sqrt{1+\dfrac{1}{n}-\dfrac{1}{{{n}^{2}}}}-2\sqrt{1-\dfrac{2}{4{{n}^{2}}}}}{1+\dfrac{3}{n}}=-1 \\ \end{align} \)