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1.
\(lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0\)
b=\(lim\left(\sqrt[3]{n^3+2n}-n+n-\sqrt{n^2+1}\right)=lim\left(\frac{2n}{\sqrt[3]{\left(n^3+2n\right)^2}+n\sqrt[3]{n^3+2n}+n^2}-\frac{1}{n+\sqrt{n^2+1}}\right)\)
\(=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0\)
c\(=lim\left(\frac{2n^2+n}{\sqrt[3]{\left(n^3+n\right)^2}+\sqrt[3]{\left(n^3+n\right)\left(n^3-2n^2\right)}+\sqrt[3]{\left(n^3-2n^2\right)^2}}\right)\)
\(=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}\)
2.
a\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)
\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)
\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)
\(\lim\limits\left(\sqrt{2n^2+3}-\sqrt{n^2+1}\right)=\lim\limits\frac{n^2-2}{\left(\sqrt{2n^2+3}+\sqrt{n^2+1}\right)}=\lim\limits\frac{n-\frac{2}{n}}{\sqrt{2+\frac{3}{n^2}}+\sqrt{1+\frac{1}{n^2}}}=+\infty\)
\(\lim\limits\frac{1}{\sqrt{n+1}-\sqrt{n}}=\lim\limits\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
a) lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
= lim \(\left(2n+1\right)^2.\frac{\left(1-\frac{1}{n}\right)}{\sqrt[3]{1+\frac{7}{n^2}-\frac{2}{n^3}}}\)
\(=+\infty\)
b) lim \(\left(2n-1\right)\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)
= lim \(\left(2-\frac{1}{n}\right)\sqrt{\frac{2+\frac{5}{n^2}}{1+\frac{1}{n^2}+\frac{2}{n^4}}}\)
=2.2 = 4
c ) = lim \(n.\frac{n^2}{\sqrt[3]{\left(n^3+n^2\right)^2+n\sqrt[3]{n^3+n^2}+n^2}}\)
= lim \(n.\frac{1}{\sqrt[3]{\left(1+\frac{1}{n}\right)^2+\sqrt[3]{1+\frac{1}{n}}+1}}\)
\(=+\infty\)
\(a=\lim\limits n\left(\sqrt[3]{\frac{1}{n}+1}+1\right)=+\infty.2=+\infty\)
\(b=\lim\limits\frac{n^2+2\sqrt{n}+3}{2n^2+n-\sqrt{n}}=\lim\limits\frac{1+\frac{2}{n\sqrt{n}}+\frac{3}{n^2}}{2+\frac{1}{n}-\frac{1}{n\sqrt{n}}}=\frac{1}{2}\)
\(c=\lim\limits\frac{2n\sqrt{n}+3}{n^2+n+1}=\frac{\frac{2}{\sqrt{n}}+\frac{3}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}=\frac{0}{1}=0\)
\(d=\lim\limits\frac{2n^2+6n\sqrt{n}}{n^2+3n+2}=\lim\limits\frac{2+\frac{6}{\sqrt{n}}}{1+\frac{3}{n}+\frac{2}{n^2}}=\frac{2}{1}=2\)
\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)
\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)
Ta có:
\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)
\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)
\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)
\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)
\(\lim\limits\dfrac{\sqrt{2\cdot4^n+1}-2^n}{\sqrt{2\cdot4^n+1}+2^n}\)
\(=\lim\limits\dfrac{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}-2^n}{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}+2^n}\)
\(=\lim\limits\dfrac{\sqrt{2+\dfrac{1}{4^n}}-1}{\sqrt{2+\dfrac{1}{4^n}}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-2\sqrt{2}}{2-1}=3-2\sqrt{2}\)
=>a=3; b=-2
\(a^3+b^3=3^3+\left(-2\right)^3=27-8=19\)