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(D) For a function $f(x)$ to be continuous at $x=0$, the condition $\lim_{x \rightarrow 0} f(x) = f(0)$ must hold. Given $f(x) = \frac{2 \sin x - \sin

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(D) For a function $f(x)$ to be continuous at $x=0$, the condition $\lim_{x ightarrow 0} f(x) = f(0)$ must hold. Given $f(x) = \frac{2 \sin x - \sin 2x}{2x \cos x}$ for $x 
eq 0$. We know that $\sin 2x = 2 \sin x \cos x$. Substituting this into the expression: $\lim_{x ightarrow 0} f(x) = \lim_{x ightarrow 0} \frac{2 \sin x - 2 \sin x \cos x}{2x \cos x}$ $= \lim_{x ightarrow 0} \frac{2 \sin x (1 - \cos x)}{2x \cos x}$ $= \lim_{x ightarrow 0} \left( \frac{\sin x}{x} ight) \cdot \left( \frac{1 - \cos x}{\cos x} ight)$ Since $\lim_{x ightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x ightarrow 0} \frac{1 - \cos x}{\cos x} = \frac{1 - 1}{1} = 0$. Therefore, $\lim_{x ightarrow 0} f(x) = 1 \cdot 0 = 0$. Since $f(0) = a$, for continuity, we must have $a = 0$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

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