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

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(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{2 - \sqrt{x + 4}}{\sin 2x}$ is in the $\frac{0}{0

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$-1/8$

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(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{2 - \sqrt{x + 4}}{\sin 2x}$ is in the $\frac{0}{0}$ indeterminate form.Applying $L'Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:$f(0) = \lim_{x 	o 0} \frac{\frac{d}{dx}(2 - \sqrt{x + 4})}{\frac{d}{dx}(\sin 2x)}$$f(0) = \lim_{x 	o 0} \frac{-\frac{1}{2\sqrt{x + 4}}}{2 \cos 2x}$Substituting $x = 0$:$f(0) = \frac{-\frac{1}{2\sqrt{0 + 4}}}{2 \cos(0)} = \frac{-\frac{1}{2(2)}}{2(1)} = \frac{-1/4}{2} = -\frac{1}{8}$.

If $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}, (x \neq 0),$ is a continuous function at $x = 0$,then $f(0)$ equals:

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