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9 tháng 4 2017

a) lim = = 2;

b) lim = = 0.

12 tháng 5 2016

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

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

 

15 tháng 8 2016

lim\(\frac{n^2+2n-3}{n\left(n+1\right)}\)=lim\(\frac{\frac{n^2}{n^2}+\frac{2n}{n^2}-\frac{3}{n^2}}{\frac{n^2}{n^2}+\frac{1}{n}}\)

=\(\frac{lim1+lin\frac{n}{2}-lim\frac{3}{n^2}}{lim1+lim\frac{1}{n}}=1\)

4 tháng 5 2016

Ta có \(\frac{x^n-nx+n-1}{\left(x-1\right)^2}=\frac{\left(x^n-1\right)-n\left(x-1\right)}{\left(x-1\right)^2}\)

                            \(=\frac{\left(x-1\right)\left(x^{n-1}+x^{n-1}+....+x+1-n\right)}{\left(x-1\right)^2}\)  (1)

Từ (1) suy ra :

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

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

         \(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left[\left(x^{n-1}+x^{n-3}+.....+x+1\right)+.....+\left(x+1\right)+1\right]}{x-1}\)

         \(=\lim\limits_{x\rightarrow1}\left[1+\left(x+1\right)+....+\left(x^{n-2}+x^{n-3}+.....+x+1\right)\right]\)

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

12 tháng 5 2016

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

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

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

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