Tìm lim biết

Question

a) $u_{n} = \frac{2n^{4} - (-1)^{n}cos^{n}n}{n^{5} + 1}$

b) $u_{n} = \frac{n^{2} + 2\sqrt{n} + 3}{2n^{2} + n - \sqrt{n}}$

c) $u_{n} = \frac{\sqrt{4n^{2} + 3n+ 1} - 2n}{\sqrt[3]{8n^{3} + 2n^{2} - 1} - 2n}$

d) $u_{n} = \sqrt{4n^{2}+ n} + \sqrt[3]{2n^{2} - 8n^{3}}$

score 1 · Answer 1

c) Xét $lim (\sqrt{4n^2+3n+1}-2n) = lim \frac{3n+1}{\sqrt{4n^2+3n+1}+2n} = lim \frac{3+\frac{1}n}{\sqrt{4+\frac{3}n+\frac{1}{n^2}}+2} = \frac{3}{4}$

$lim (\sqrt[3]{8n^3+2n-1} -2n)= lim \frac{2n-1}{\sqrt[3]{(8n^3+2n-1)^2}+2n.\sqrt[3]{8n^3+2n-1}+4n^2}= lim \frac{\frac{2}n - \frac{1}{n^2}}{\sqrt[3]{(8+\frac{2}{n^2}-\frac{1}{n^3})^2}+2.\sqrt[3]{8+\frac{2}{n^2}-\frac{1}{n^3}}+4}=\frac{0}{\sqrt[3]{64}+2.\sqrt[3]{8}+4}=0$

vì $n \to + \infty \Rightarrow \frac{2}{n}-\frac{1}{n^2} >0 $

$\Rightarrow lim \frac{\frac{2}n - \frac{1}{n^2}}{\sqrt[3]{(8+\frac{2}{n^2}-\frac{1}{n^3})^2}+2.\sqrt[3]{8+\frac{2}{n^2}-\frac{1}{n^3}}+4} = 0^+$

$\Rightarrow lim \sqrt[3]{8n^3+2n-1}-2n = 0^+$

$lim u_n=\frac{\frac{3}4}{0^+} = + \infty$