Bài 2: Giới hạn của hàm số

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 "

\(L=\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_1\right)......\left(x+a_1\right)}-x\right)=\frac{a_1+a_2+....+a_n}{n}\)

Ta có \(L_m=\lim\limits_{x\rightarrow1}\left(\frac{m-\left(1+x+x^2+.....+x^{m-1}\right)}{1-x^m}\right)\)

               \(=\lim\limits_{x\rightarrow1}\frac{\left(1-x\right)+\left(1-x^2\right)+.....+\left(1-x^{m-1}\right)}{1-x^m}\)

               \(=\lim\limits_{x\rightarrow1}\frac{\left(1-x\right)\left[1+\left(1+x\right)+.....+\left(1+x+x^2+.....+x^{m-2}\right)\right]}{\left(1-x\right)\left(1+x+x^2+.....+x^{m-1}\right)}\)

               \(=\frac{1+2+3+....+\left(m-1\right)}{m}=\frac{\left(m-1\right)m}{2m}=\frac{m-1}{2}\)

1. Ta có : \(\lim\limits_{x\rightarrow0}\frac{\tan ax}{\tan bx}=\lim\limits_{x\rightarrow0}\left(\frac{\sin ax}{\sin bx}.\frac{\cos ax}{\cos bx}\right)=\lim\limits_{x\rightarrow0}\frac{\sin ax}{\sin bx}=\lim\limits_{x\rightarrow0}\left(\frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}}.\frac{ax}{bx}\right)=\frac{a}{b}\frac{\lim\limits_{x\rightarrow0}\frac{\sin ax}{ax}}{\lim\limits_{x\rightarrow0}\frac{\sin bx}{bx}}=\frac{a}{b}\frac{\lim\limits_{y\rightarrow0}\frac{\sin y}{y}}{\lim\limits_{z\rightarrow0}\frac{\sin z}{z}}=\frac{a}{b}\)

2. Ta có : \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\sin^2\frac{ax}{2}}{x^2}=\lim\limits_{x\rightarrow0}\left[\left(\frac{\sin\frac{ax}{2}.\sin\frac{ax}{2}}{\frac{ax}{2}.\frac{ax}{2}}\right).\frac{a^2}{2}\right]\)

                                   \(=\frac{a^2}{2}\left(\lim\limits_{y\rightarrow0}\frac{\sin y}{y}\right)^2=\frac{a^2}{2}\)

Ta có

\(lim_{x-->x0}=\frac{1}{\sqrt[2]{\left(0+1\right)^2+\sqrt[2]{0+1}+1}}=\frac{1}{\sqrt[2]{1^2+\sqrt[2]{1}+1}}=\frac{1}{\sqrt[2]{4}}=\frac{1}{2}\)