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(D) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0)$.Given $f(0) = a$.Now,calculate the limit: $\lim_{x \right

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(D) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0)$.Given $f(0) = a$.Now,calculate the limit: $\lim_{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{2x \cos x}$.Using the identity $\sin 2x = 2 \sin x \cos x$,we get:$\lim_{x \rightarrow 0} \frac{2 \sin x - 2 \sin x \cos x}{2x \cos x} = \lim_{x \rightarrow 0} \frac{2 \sin x (1 - \cos x)}{2x \cos x}$.$= \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right) \cdot \left( \frac{1 - \cos x}{\cos x} \right)$.Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{\cos x} = \frac{1 - 1}{1} = 0$.Therefore,$\lim_{x \rightarrow 0} f(x) = 1 \cdot 0 = 0$.Since $f(x)$ is continuous at $x = 0$,$a = 0$.

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

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