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Question

a) $lim\frac{\sqrt{4n^2+1}-(2n+1)}{\sqrt{n^2+4n+1}-n}$

b) $lim\frac{\sqrt[3]{2-n^3}+n}{\sqrt{n^2+1}-n}$

c) $lim(\sqrt[3]{n^3-3n^2+1}-\sqrt{n^2+4n})$

score 1 · Answer 1

b. Ta có:

$\lim\dfrac{\sqrt[3]{2-n^3}+n}{\sqrt{n^2+1}-n}$

$=\lim(n-\sqrt[3]{n^3-2})(\sqrt{n^2+1}+n)$

$=\lim\dfrac{2(\sqrt{n^2+1}+n)}{n^2+n\sqrt[3]{n^3-2}+\sqrt[3]{(n^3-2)^2}}$

$=\lim\dfrac{2(\sqrt{1+\dfrac{1}{n^2}}+1)}{n+\sqrt[3]{n^3-2}+\sqrt[3]{n^3(1-\dfrac{2}{n^3})^2}}=0$

score 1 · Answer 2

c)

$lim(\sqrt[3]{n^3-3n^2+1}-\sqrt{n^2+4n})$

$=lim(\sqrt[3]{n^3-3n^2+1}-n+n-\sqrt{n^2+4n})$

$=lim(\sqrt[3]{n^3-3n^2+1}-n)+lim(n-\sqrt{n^2+4n})$

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

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

$=-1-2=-3$

score 1 · Answer 3

a)

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

$=lim\frac{-4}{4+\frac{1}{n}}\times \frac{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}{\sqrt{4+\frac{1}{n^2}}+2+\frac{1}{n}}$

$=-\frac{1}{2}$

3 Đáp án