Question

[Nguyễn Trọng Minh Tín]

Tính giới hạn hàm số :

\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\tan x}\)

 

Answer 1

[Trần Phan Ngọc Hân]

\(L=\lim\limits_{x\rightarrow0}\frac{\ln x-1}{\tan x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\frac{\sin x}{\cos x}}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{2x.\frac{\sin x}{x}.\frac{1}{2\cos x}}\)

   \(=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+2x\right)}{2x}.\frac{1}{\frac{\sin x}{x}}.2\cos x\right]=1.\frac{1}{1}.2.1=2\)