If the function [f(x)=( 3 x+ x- 3 x)/(x^3), x R], is continuous at x=0, then f(0) is equal…

Question

If the function $$f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}$$, is continuous at $x=0$, then $f(0)$ is equal to :

Accepted Answer

$$\begin{aligned}
& \lim _\limits{x \rightarrow 0} f(x)=f(0) \quad \text { (continuous at } x=0) \
& \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}
\end{aligned}$$

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 $$\alpha+3=0 \Rightarrow \alpha=-3$$

$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)$$

Solutions · 1