If $f\left(x\right)= \begin{cases} \dfrac{\sin 3x}{7x} + \alpha, & x > 0 \\ x+3-\beta, & x < 0 \end{cases}$ $\;\;$ is continuous at $x=0$, find $\alpha + \beta$
Since $f\left(x\right)$ is continuous at $x=0$,
$\lim\limits_{x \to 0^-} f\left(x\right) = \lim\limits_{x \to 0^+}f\left(x\right)$
i.e. $\lim\limits_{x \to 0^-} \; \left(x+3-\beta\right) = \lim\limits_{x \to 0^+} \; \left(\dfrac{\sin 3x}{7x}+\alpha\right)$
i.e. $3-\beta = \dfrac{3}{7} \; \lim\limits_{x \to 0^+} \; \dfrac{\sin 3x}{3x}+ \lim\limits_{x \to 0^+} \alpha$
i.e. $3-\beta = \dfrac{3}{7} \times 1 + \alpha$
i.e. $\alpha + \beta = 3-\dfrac{3}{7}=\dfrac{18}{7}$