Question

$\underset{x\to 0}{\lim }\frac{2x-\sin x}{\tan x+x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{2x-\sin x}{\tan x+x}\right]$

Dividing the numerator and the denominator by x, we get:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{2-{\displaystyle \frac{\sin x}{x}}}{{\displaystyle \frac{\tan x}{x}}+1}\right] \\ =\frac{2-1}{1+1}\left[\because \underset{x\to 0}{\lim }\frac{\tan x}{x}=1,\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right] \\ =\frac{1}{2} \end{gathered}$$