HOC24
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Môn học
Chủ đề / Chương
Bài học
C
\(=lim\left(\dfrac{\left(n+1\right)^2-\left(n^2+2n+5\right)}{n+1+\sqrt{n^2+2n+5}}\right)\)
\(=lim\left(\dfrac{n^2+1-n^2-2n-5}{n+1+\sqrt{n^2+2n+5}}\right)\)
\(=lim\dfrac{-2n-4}{n+1+\sqrt{n^2+2n+5}}\)
\(=lim\dfrac{\dfrac{-2n}{n}-\dfrac{4}{n}}{\dfrac{n}{n}+\dfrac{1}{n}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{2n}{n^2}+\dfrac{5}{n^2}}}\)
\(=\dfrac{-2-0}{1+0+\sqrt{1+0+0}}\)
\(=\dfrac{-2}{2}\)
\(=-1\)
\(12,lim\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}\)
\(=lim\dfrac{n^2.\sqrt[3]{\dfrac{n^6}{n^6}-\dfrac{7n^3}{n^6}-\dfrac{5n}{n^6}+\dfrac{8}{n^6}}}{n^2\left(\dfrac{n}{n^2}+\dfrac{12}{n^2}\right)}\)
\(=lim\dfrac{n^2.\sqrt[3]{1-\dfrac{7}{n^3}-\dfrac{5}{n^5}+\dfrac{8}{n^6}}}{n^2\left(\dfrac{1}{n}-\dfrac{12}{n^2}\right)}\)
\(=+\infty\)
vì \(\left\{{}\begin{matrix}lim n^2\left(\sqrt{1-\dfrac{7}{n^3}-\dfrac{5}{n^5}+\dfrac{8}{n^6}}\right)=1>0\\lim n^2\left(\dfrac{1}{n}+\dfrac{12}{n^2}\right)=0\end{matrix}\right.\)
12,