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Lời giải:
\(C=\lim\limits_{x\to +\infty}\left[x\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})}-x\right]\)
\(=\lim\limits_{x\to +\infty}x\left[\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1\right]\)
\(=\lim\limits _{x\to +\infty}\frac{\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1}{(1+\frac{a_1}{x})(1+\frac{a_2}{x})..(1+\frac{a_n}{x})-1}.\frac{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}(A.B)=\lim\limits_{x\to +\infty}A.\lim\limits_{x\to +\infty}B\)
Với $A$. Đặt \(\sqrt[n]{\prod_{i=1}^n (1+\frac{a_i}{x})}=u\). \(x\to +\infty\Rightarrow \frac{a_i}{x}\to 0\Rightarrow 1+\frac{a_i}{x}\to 1\Rightarrow u\to 1\)
\(\lim\limits_{x\to +\infty}A=\lim\limits_{u\to 1}\frac{u-1}{u^n-1}=\lim\limits_{u\to 1}\frac{1}{u^{n-1}+...+1}=\frac{1}{n}\)
Với $B$
\(\lim\limits _{x\to +\infty}B=\lim\limits _{x\to +\infty}\frac{1+\frac{a_1+a_2+..+a_n}{x}+\frac{a_1a_2+a_2a_3+...+a_{n-1}a_n}{x^2}+....-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}\left(a_1+a_2+...+a_n+\frac{a_1a_2+...+a_{n-1}a_n}{x}+...\right)=a_1+a_2+..+a_n\)
Do đó: $C=\frac{a_1+a_2+...+a_n}{n}$
Đáp án C
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
\(\lim\limits_{x\rightarrow+\infty}\left(x-3\right)\left(\sqrt{x^2+x+3}-\sqrt{x^2+x+1}\right)\)
=\(\lim\limits_{x\rightarrow+\infty}\left(x-3\right)\frac{x^2+x+3-x^2-x-1}{\sqrt{x^2+x+3}+\sqrt{x^2+x+1}}\)
=\(\lim\limits_{x\rightarrow+\infty}\left(x-3\right)\frac{2}{\sqrt{x^2+x+3}+\sqrt{x^2+x+1}}\)
=\(\lim\limits_{x\rightarrow+\infty}\frac{2\left(1-\frac{3}{x}\right)}{\sqrt{1+\frac{1}{x}+\frac{3}{x^2}}+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}}=1\)
Bài 1:
\(\lim\limits _{x\to 1}\frac{4x^6-5x^5+x}{(1-x)^2}=\lim\limits _{x\to 1}\frac{x(x-1)^2(4x^3+3x^2+2x+1)}{(1-x)^2}\)
\(=\lim\limits _{x\to 1}x(4x^3+3x^2+2x+1)=1(4.1^3+3.1^2+2.1+1)=10\)
Bài 3:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-(ax+b)]=0\)
\(\Rightarrow \lim\limits _{x\to +\infty}\frac{\sqrt{9x^2-4x+3}-(ax+b)}{x}=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\left(\sqrt{9-\frac{4}{x}+\frac{3}{x^2}}-a+\frac{b}{x}\right)=0\)
\(\Leftrightarrow a=3\)
Thay $a=3$ vào đk ban đầu:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-3x-b]=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty} (\sqrt{9x^2-4x+3}-3x)=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4x+3}{\sqrt{9x^2-4x+3}+3x}=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4+\frac{3}{x}}{\sqrt{9-\frac{4}{x}+\frac{3}{x}}+3}=b\)
\(\Leftrightarrow \frac{-4}{6}=b\Leftrightarrow b=-\frac{2}{3}\)
Giới hạn trên có dạng \(\infty-\infty\), ta đưa nó về dạng \(\frac{0}{0}\) nhờ phép biến đổi sau :
Đặt \(x=\frac{1}{y}\), khi \(x\rightarrow+\infty\) thì \(y\rightarrow0\)
Ta có : \(L=\lim\limits_{y\rightarrow0}\frac{\sqrt[3]{\left(1+a_1y\right)\left(1+a_2y\right)\left(1+a_3y\right)}-1}{y}\)
Áp dụng phép đổi biến \(x=\frac{1}{y}\) ta có "
\(L=\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_1\right)......\left(x+a_1\right)}-x\right)=\frac{a_1+a_2+....+a_n}{n}\)