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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 :
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The correct answer is:
-4
$f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}$
is continuous at $\mathrm{x}=0$
$\lim _{x \rightarrow 0}=\frac{3 x-\frac{(3 x)^3}{\underline{|3}}+\ldots+\alpha\left(x-\frac{x^3}{\underline{|3}} \ldots\right)-\beta\left(1-\frac{(3 x)^2}{\lfloor 2} \ldots\right)}{x^3}=f(0)$
$\begin{aligned}
& \lim _{x \rightarrow 0}= \frac{-\beta+x(3+\alpha)+\frac{9 \beta x^2}{\underline{|2}}+\left(\frac{-27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}\right) x^3 \ldots}{x^3}=f(0) \\
& \text { for exist } \\
& \beta=0,3+\alpha=0,-\frac{27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}=f(0)
\end{aligned}$
$\begin{aligned} & \alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0) \\ & f(0)=\frac{-27+3}{6}=-4\end{aligned}$
is continuous at $\mathrm{x}=0$
$\lim _{x \rightarrow 0}=\frac{3 x-\frac{(3 x)^3}{\underline{|3}}+\ldots+\alpha\left(x-\frac{x^3}{\underline{|3}} \ldots\right)-\beta\left(1-\frac{(3 x)^2}{\lfloor 2} \ldots\right)}{x^3}=f(0)$
$\begin{aligned}
& \lim _{x \rightarrow 0}= \frac{-\beta+x(3+\alpha)+\frac{9 \beta x^2}{\underline{|2}}+\left(\frac{-27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}\right) x^3 \ldots}{x^3}=f(0) \\
& \text { for exist } \\
& \beta=0,3+\alpha=0,-\frac{27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}=f(0)
\end{aligned}$
$\begin{aligned} & \alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0) \\ & f(0)=\frac{-27+3}{6}=-4\end{aligned}$
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