Đặt \(A=\frac{1}{\sqrt{n^3+1}}+\frac{1}{\sqrt{n^3+2}}+...+\frac{1}{\sqrt{n^3+n}}\)
\(n^3+1< n^3+2< ...< n^3+n\)
\(\Rightarrow\frac{1}{\sqrt{n^3+1}}>\frac{1}{\sqrt{n^3+2}}>...>\frac{1}{\sqrt{n^3+n}}\)
\(\Rightarrow\frac{n}{\sqrt{n^3+n}}< A< \frac{n}{\sqrt{n^3+1}}\)
Mà \(lim\left(\frac{n}{\sqrt{n^3+1}}\right)=lim\left(\frac{n}{\sqrt{n^3+n}}\right)=0\)
\(\Rightarrow lim\left(A\right)=0\)
cho f(n) = \(\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{n}}\) nϵN*. GIá trị lim\(\frac{f\left(n\right)}{n^2+1}\) bằng ?
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
a)lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
b)lim [(2n-1)\(\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)]
c)lim [n(\(\sqrt[3]{n^3+n^2}-n\))]
\\(\\lim\\limits_{x\\rightarrow8}\\frac{\\sqrt[3]{x}-2}{2x-16}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow-2}\\frac{\\sqrt{x-3}-1}{\\sqrt[3]{x-6}+2}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow1}\\frac{2x-1-\\sqrt{x^2+2x-2}}{x^2-4x+3}\\)
\nTìm \(\lim\limits_{x->-\infty}\)\(\frac{\left|x\right|\sqrt{4x^2+3}}{2x-1}\)
lim \(\sqrt{n}\)(\(\sqrt{n+4}\)-\(\sqrt{n+3}\))
lim (n-2-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)
\(lim\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+......+\frac{1}{\sqrt{n^2+n}}\right)\)
\(\lim\limits\frac{\sqrt{n}+\sqrt[3]{n}+\sqrt[4]{n}}{\sqrt{2n+1}}\)
Với n là số nguyên dương, đặt \(S_n=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}\). Khi đó \(limS_n\) bằng ?
tính các giới hạn sau:
a, lim\(\frac{n^{2020}-n+1}{n^{2022}+2n-3}\)
b, lim(\(\sqrt[3]{n^3-2n^2}-n\))
c, lim \(\left(\sqrt{n^2+3n}-n+2\right)\)
d, lim \(n\left(\sqrt{n^2-1}-\sqrt{n^2+2}\right)\)
