Tính các giới hạn:
a) \(lim \frac{(n+1)(3-2n)^2}{n^3+1}\)
b) \(lim \left ( \frac{1}{n^2+1}+\frac{2}{n^2+1}+\frac{3}{n^2+1}+...+ \frac{n-1}{n^2+1} \right );\)
c) \(lim \frac{\sqrt{4n^2+1}+n}{2n+1}\)
d) \(lim \sqrt{n}(\sqrt{n-1}-\sqrt{n})\)
Câu a:
\(lim \frac{(n+1)(3-2n)^2}{n^3+1}=lim \frac{\left ( 1+\frac{1}{n} \right )\left ( \frac{3}{n-2} \right )^2}{1+\frac{1}{n^3}}=4.\)
Câu b:
\(lim \left ( \frac{1}{n^2+1}+\frac{2}{n^2+1}+\frac{3}{n^2+1}+...+ \frac{n-1}{n^2+1} \right )\)
\(= lim\frac{1+2+3+...+(n-1)}{n^2+1}\)
\(= lim \frac{(n-1)n}{2(n^2+1)}=\frac{1}{2}\)
Câu c:
\(lim \frac{\sqrt{4n^2+1}+n}{2n+1} =lim\frac{\sqrt{4+\frac{1}{n^2}}+1}{2+\frac{1}{n}}=\frac{3}{2}\)
Câu d:
\(lim \sqrt{n}(\sqrt{n-1}-\sqrt{n})= lim\frac{\sqrt{n}(n-1)-n}{\sqrt{n-1}+\sqrt{n}}\)
\(=lim \frac{-\sqrt{n}}{\sqrt{n-1}+\sqrt{n}}= lim \frac{-1}{\sqrt{1-\frac{1}{n}}+1}=-\frac{1}{2}\)
-- Mod Toán 11
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