Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{x^3}$ is continuous at $\mathrm{x}=0$ ${\lim }_{x\to 0}=\frac{3x-\frac{(3x{)}^3}{\underline{|3}}+\dots +\alpha \left(x-\frac{x^3}{\underline{|3}}\dots \right)-\beta \left(1-\frac{(3x{)}^2}{\lfloor 2}\dots \right)}{x^3}=f(0)$ $\begin{array}{rl}& \underset{x\to 0}{\lim }=\frac{-\beta +x(3+\alpha )+\frac{9\beta x^2}{\underline{|2}}+\left(\frac{-27}{\underline{|3}}-\frac{\alpha }{\underline{|3}}\right)x^3\dots }{x^3}=f(0)\\ & \text{ for exist }\\ & \beta =0,3+\alpha =0,-\frac{27}{\underline{|3}}-\frac{\alpha }{\underline{|3}}=f(0)\end{array}$ $\begin{array}{rl}& \alpha =-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0)\\ & f(0)=\frac{-27+3}{6}=-4\end{array}$