If f(x) = begin{cases} frac{sin((p+1)x) + sin | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First,calculate the Right

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$(-3/2, 1/2)$

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First,calculate the Right Hand Limit $(RHL)$:$RHL = \lim_{x 	o 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \cdot \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{1+x} - 1}{x}$.Multiplying by the conjugate $\frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}$,we get:$RHL = \lim_{x 	o 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x 	o 0^+} \frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{1+1} = \frac{1}{2}$.Next,calculate the Left Hand Limit $(LHL)$:$LHL = \lim_{x 	o 0^-} \frac{\sin((p+1)x) + \sin x}{x} = \lim_{x 	o 0^-} \left( \frac{\sin((p+1)x)}{x} + \frac{\sin x}{x} ight) = (p+1) + 1 = p+2$.Since $f(0) = q$,for continuity,we require $LHL = RHL = f(0)$:$p + 2 = 1/2 \implies p = 1/2 - 2 = -3/2$.$q = 1/2$.Thus,the ordered pair $(p, q) = (-3/2, 1/2)$.

If $f(x) = \begin{cases} \frac{\sin((p+1)x) + \sin x}{x} & , x < 0 \\ q & , x = 0 \\ \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} & , x > 0 \end{cases}$ is continuous at $x = 0$,then the ordered pair $(p, q)$ is equal to

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