If the function f(x) = frac{sin 3x + alpha si | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Using the Taylor series expa

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$-4$

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Using the Taylor series expansions:$\sin 3x = 3x - \frac{(3x)^3}{3!} + \dots = 3x - \frac{27x^3}{6} + \dots$$\sin x = x - \frac{x^3}{6} + \dots$$\cos 3x = 1 - \frac{(3x)^2}{2!} + \dots = 1 - \frac{9x^2}{2} + \dots$Substituting these into the expression for $f(x)$:$f(x) = \frac{(3x - \frac{27x^3}{6} + \dots) + \alpha(x - \frac{x^3}{6} + \dots) - \beta(1 - \frac{9x^2}{2} + \dots)}{x^3}$$f(x) = \frac{-\beta + x(3 + \alpha) + x^2(\frac{9\beta}{2}) + x^3(-\frac{27}{6} - \frac{\alpha}{6}) + \dots}{x^3}$For the limit to exist as $x 	o 0$,the coefficients of $x^0$,$x^1$,and $x^2$ must be zero:$1$) $-\beta = 0 \implies \beta = 0$$2$) $3 + \alpha = 0 \implies \alpha = -3$$3$) $\frac{9\beta}{2} = 0$ (which is satisfied since $\beta = 0$)Now,the limit is the coefficient of $x^3$:$f(0) = -\frac{27}{6} - \frac{\alpha}{6} = \frac{-27 - (-3)}{6} = \frac{-24}{6} = -4$.

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

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