Question

$\underset{x\to 0}{\lim }\frac{5x\cos x+3\sin x}{3x^2+\tan x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{5x\cos x+3\sin x}{3x^2+\tan x}\right]$

Dividing the numerator and the denominator by x:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{5\cos x+3\left({\displaystyle \frac{\sin x}{x}}\right)}{3x+\left({\displaystyle \frac{\tan x}{x}}\right)}\right] \\ =\left[\frac{5\cos 0+3}{3\times 0+1}\right]\left[\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1,\underset{x\to 0}{\lim }\frac{\tan x}{x}=1\right] \\ =\frac{5+3}{0+1} \\ =8 \end{gathered}$$