Câu hỏi của Thu Diện

Question

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Lê Song Phương · Accepted Answer

a) $\lim\limits_{ }\left(\sqrt{n^2-n+1}-n\right)$

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

$=\lim\limits_{ }\left(\dfrac{1-n}{\sqrt{n^2-n+1}+n}\right)$

$=\lim\limits_{ }\left(\dfrac{\dfrac{1}{n}-1}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)$

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

b) $\lim\limits_{ }\left(\dfrac{-3}{4n^2-2n+1}\right)=0$

c) $\lim\limits_{ }\dfrac{n^2+n+5}{2n+1}=+\infty$

d) $\lim\limits_{ }\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)$

$=\lim\limits_{ }\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)$

$\lim\limits_{ }\left(\dfrac{-2n-\dfrac{3}{n}}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)$

$=-\infty$Câu trả lời: a) $\lim\limits_{ }\left(\sqrt{n^2-n+1}-n\right)$

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

$=\lim\limits_{ }\left(\dfrac{1-n}{\sqrt{n^2-n+1}+n}\right)$

$=\lim\limits_{ }\left(\dfrac{\dfrac{1}{n}-1}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)$

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

b) $\lim\limits_{ }\left(\dfrac{-3}{4n^2-2n+1}\right)=0$

c) $\lim\limits_{ }\dfrac{n^2+n+5}{2n+1}=+\infty$

d) $\lim\limits_{ }\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)$

$=\lim\limits_{ }\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)$

$\lim\limits_{ }\left(\dfrac{-2n-\dfrac{3}{n}}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)$

$=-\infty$

Nguyễn Lê Phước Thịnh · Answer

a: $lim\left(\sqrt{n^2-n+1}-n\right)$$=\lim\limits\dfrac{n^2-n+1-n^2}{\sqrt{n^2-n+1}+n}=\lim\limits\dfrac{-n+1}{\sqrt{n^2-n+1}+n}$$=\lim\limits\dfrac{-1+\dfrac{1}{n}}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{-1+0}{\sqrt{1-0+0}+1}=\dfrac{-1}{2}$b: $\lim\limits\dfrac{-3}{4n^2-2n+1}$$=\lim\limits\dfrac{-\dfrac{3}{n^2}}{4-\dfrac{2}{n}+\dfrac{1}{n^2}}=\dfrac{0}{4-0+0}=0$c: $\lim\limits\dfrac{n^2+n+5}{2n+1}=\lim\limits\dfrac{n^2\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{n\left(2+\dfrac{1}{n}\right)}$$=\lim\limits\dfrac{n\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{2+\dfrac{1}{n}}=+\infty$d: $\lim\limits\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)$$=\lim\limits\left(\dfrac{n^2-1-3n^2-2}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)=\lim\limits\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)$$=\lim\limits\left(\dfrac{n^2\left(-2-\dfrac{3}{n^2}\right)}{n\cdot\left(\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}\right)}\right)$$=\lim\limits\left(\dfrac{n\left(-2-\dfrac{3}{n^2}\right)}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)=+\infty$Câu trả lời: a: $lim\left(\sqrt{n^2-n+1}-n\right)$$=\lim\limits\dfrac{n^2-n+1-n^2}{\sqrt{n^2-n+1}+n}=\lim\limits\dfrac{-n+1}{\sqrt{n^2-n+1}+n}$$=\lim\limits\dfrac{-1+\dfrac{1}{n}}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{-1+0}{\sqrt{1-0+0}+1}=\dfrac{-1}{2}$b: $\lim\limits\dfrac{-3}{4n^2-2n+1}$$=\lim\limits\dfrac{-\dfrac{3}{n^2}}{4-\dfrac{2}{n}+\dfrac{1}{n^2}}=\dfrac{0}{4-0+0}=0$c: $\lim\limits\dfrac{n^2+n+5}{2n+1}=\lim\limits\dfrac{n^2\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{n\left(2+\dfrac{1}{n}\right)}$$=\lim\limits\dfrac{n\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{2+\dfrac{1}{n}}=+\infty$d: $\lim\limits\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)$$=\lim\limits\left(\dfrac{n^2-1-3n^2-2}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)=\lim\limits\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)$$=\lim\limits\left(\dfrac{n^2\left(-2-\dfrac{3}{n^2}\right)}{n\cdot\left(\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}\right)}\right)$$=\lim\limits\left(\dfrac{n\left(-2-\dfrac{3}{n^2}\right)}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)=+\infty$

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