[Toán 11] Giới hạn

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niemkieuloveahbu

[TEX]\Large{\lim\frac{\sqrt{4n^2+1}-2n-1}{\sqrt{n^2+4n+1}-n}=\lim \frac{-4n(\sqrt{n^2+4n+1}+n)}{(4n+1)(\sqrt{4n^2+1}+2n+1)}=\lim \frac{-4\(\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+n)}{(4+\frac{1}{n})(\sqrt{4+\frac{1}{n^2}}+2+\frac{1}{n})}= -\frac{1}{2}\\ \lim \frac{n^2+\sqrt[3]{1-n^6}}{\sqrt{n^4+1}-n^2}=\lim \frac{\sqrt{n^4+1}+n^2}{n^4-n^2.\sqrt[3]{1-n^6}+\sqrt[3]{(1-n^6)^2}}=\lim \frac{\sqrt{1+\frac{1}{n^4}}+1}{n^2-\sqrt[3]{1-n^6}+\sqrt[3]{(\frac{1}{n^6}-1)^2}}=0[/TEX]
 
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