Question

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

Answer 1

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

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

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\cos x+{\displaystyle \frac{2\sin x}{x}}}{x+{\displaystyle \frac{\tan x}{x}}}\right] \\ =\frac{\cos 0+2\times 1}{0+1}\left[\because \underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1,\underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right)=1\right] \\ =\frac{3}{1} \end{gathered}$$