QJEEM24Mafc02: JEE Main 2024 Mathematics - Class 12 JEEM24Mafc02 Solution: If the function , is continuous at , then is equal to :

$$\begin{array}{rl}& \underset{x\to 0}{lim}f(x)=f(0)\text{ (continuous at }x=0)\\ & \underset{x\to 0}{lim}\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{x^3}\end{array}$$

For limit to exist $\beta=0$

$\begin{aligned}& \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\& \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3}\end{aligned}$

For limit to exist $\alpha+3=0 \Rightarrow \alpha=-3$

$$\Rightarrow \underset{x\to 0}{lim}\frac{-4{\sin }^{3}x}{x^3}=-4=f(0)$$