Question

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

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sqrt{x}\cos x-2\sin x}{3x+\tan x} \\ \mathrm{By}\mathrm{dividing}\mathrm{number}\mathrm{and}\mathrm{denominator}\mathrm{by}x, \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{5\cos x-2{\displaystyle \frac{\sin x}{x}}}{3+{\displaystyle \frac{\tan x}{x}}} \\ \mathrm{i}.\mathrm{e}.\frac{5\left(\begin{array}{c}\lim \\ x\to 0\end{array}\cos x\right)-2\left(\begin{array}{c}\lim \\ x\to 0\end{array}{\displaystyle \frac{\sin x}{x}}\right)}{3+\left(\begin{array}{c}\lim \\ x\to 0\end{array}{\displaystyle \frac{\tan x}{x}}\right)} \\ \mathrm{i}.\mathrm{e}.\frac{5\times 1-2\times 1}{3+1}=\frac{5-2}{4}=\frac{3}{4} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{5x\cos x-2\sin x}{3x+\tan x}=\frac{3}{4} \end{gathered}$$