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\(\lim\dfrac{1+a+...+a^n}{1+b+...+b^n}=\lim\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}=\lim\dfrac{\left(1-a^n\right)\left(1-b\right)}{\left(1-b^n\right)\left(1-a\right)}=\dfrac{1-b}{1-a}\)
\(\Rightarrow\dfrac{1-b}{1-a}=\dfrac{2}{3}\Leftrightarrow3-3b=2-2a\)
\(\Leftrightarrow2a-3b=-1\)
\(x^2+2x-3=0\) có nghiệm \(x=1\) nên giới hạn đã cho hữu hạn khi \(2x^2+ax+b=0\) cũng có nghiệm \(x=1\)
\(\Rightarrow2.1^2+a.1+b=0\Rightarrow a+b+2=0\Rightarrow b=-a-2\)
Thay vào:
\(\lim\limits_{x\rightarrow1}\dfrac{2x^2+ax-a-2}{x^2+2x-3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x+2\right)+a\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x+2+a\right)}{\left(x-1\right)\left(x+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2x+2+a}{x+3}=\dfrac{4+a}{4}=\dfrac{3}{4}\)
\(\Rightarrow4+a=3\Rightarrow a=-1\Rightarrow b=-a-2=-1\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
Nếu \(a\ne0\Rightarrow\lim\dfrac{an^3+bn^2+2n+4}{n^2+1}=\lim\dfrac{an+b+\dfrac{2}{n}+\dfrac{4}{n^2}}{1+\dfrac{1}{n}}=\infty\) ko thỏa mãn
\(\Rightarrow a=0\)
Khi đó: \(\lim\dfrac{bn^2+2n+4}{n^2+1}=\lim\dfrac{b+\dfrac{2}{n}+\dfrac{4}{n^2}}{1+\dfrac{1}{n^2}}=b\Rightarrow b=1\)
\(\Rightarrow2a+b=1\)