Let f(x) = frac{2 - sqrt{x + 4}}{sin 2x},x ne | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \rightarrow 0} f(x)$.Evaluating the limit: $\lim_{x \rightarrow 0} \frac{2 - \

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$\frac{-1}{8}$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x ightarrow 0} f(x)$.Evaluating the limit: $\lim_{x ightarrow 0} \frac{2 - \sqrt{x + 4}}{\sin 2x}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'	ext{Hospital's Rule}$:$\lim_{x ightarrow 0} \frac{\frac{d}{dx}(2 - \sqrt{x + 4})}{\frac{d}{dx}(\sin 2x)} = \lim_{x ightarrow 0} \frac{-\frac{1}{2\sqrt{x + 4}}}{2 \cos 2x}$.Substituting $x = 0$:$\frac{-\frac{1}{2\sqrt{4}}}{2 \cos(0)} = \frac{-\frac{1}{4}}{2(1)} = -\frac{1}{8}$.Thus,$f(0) = -\frac{1}{8}$.

Let $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}$,$x \neq 0$. In order for $f(x)$ to be continuous at $x = 0$,$f(0)$ must be defined as:

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