If a function f(x) = begin{cases} frac{tan (a | vedclass

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(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0^{-}} f(x) = f(0) = \beta$. First,evaluat

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$30$

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(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0^{-}} f(x) = f(0) = \beta$. First,evaluate the right-hand limit: $\lim_{x 	o 0^{+}} \frac{	an (\alpha + 1)x + 	an 2x}{x} = \lim_{x 	o 0^{+}} \frac{	an (\alpha + 1)x}{x} + \lim_{x 	o 0^{+}} \frac{	an 2x}{x} = (\alpha + 1) + 2 = \alpha + 3$. Thus,$\beta = \alpha + 3$. Next,evaluate the left-hand limit: $\lim_{x 	o 0^{-}} \frac{\sin 3x - 	an 3x}{x^{3}}$. Using the Taylor series expansion $\sin 3x = 3x - \frac{(3x)^3}{6} + O(x^5)$ and $	an 3x = 3x + \frac{(3x)^3}{3} + O(x^5)$,we get: $\lim_{x 	o 0^{-}} \frac{(3x - \frac{27x^3}{6}) - (3x + \frac{27x^3}{3})}{x^3} = \lim_{x 	o 0^{-}} \frac{-\frac{9}{2}x^3 - 9x^3}{x^3} = -\frac{9}{2} - 9 = -\frac{27}{2}$. So,$\beta = -\frac{27}{2}$. Substituting $\beta$ into the first equation: $-\frac{27}{2} = \alpha + 3 \implies \alpha = -\frac{27}{2} - 3 = -\frac{33}{2}$. Finally,$|\alpha| + |\beta| = |-\frac{33}{2}| + |-\frac{27}{2}| = \frac{33}{2} + \frac{27}{2} = \frac{60}{2} = 30$.

If a function $f(x) = \begin{cases} \frac{\tan (\alpha + 1)x + \tan 2x}{x}, & \text{if } x > 0 \\ \beta, & \text{at } x = 0 \\ \frac{\sin 3x - \tan 3x}{x^{3}}, & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $|\alpha| + |\beta| =$

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