Question

$\underset{x\to 0}{\lim }\frac{7x\cos x-3\sin x}{4x+\tan x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{7x\cos x-3\sin x}{4x+\tan x}\right]$

It is of the form $\left(\frac{0}{0}\right)$.
Dividing the numerator and the denominator by x:

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\frac{7\cos x-{\displaystyle 3\left(\frac{\sin x}{x}\right)}}{4+{\displaystyle \left(\frac{\tan x}{x}\right)}} \\ \Rightarrow \frac{7\underset{x\to 0}{\lim }\left(\cos x\right)-{\displaystyle 3\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)}}{4+{\displaystyle \underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right)}} \\ \Rightarrow \frac{7.1-3.1}{4+1} \\ \Rightarrow \frac{4}{5} \end{gathered}$$