The values of p and q for which the function | vedclass

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

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

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(A) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = q$.First,calculate the Right Hand Limit ($R$.$H$.$L$.):$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \cdot \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{1+x} - 1}{x}$.Multiplying by the conjugate $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$,we get:$\lim_{x 	o 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x 	o 0^+} \frac{x}{x(\sqrt{1+x} + 1)} = \lim_{x 	o 0^+} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{1+1} = \frac{1}{2}$.Thus,$q = 1/2$.Next,calculate the Left Hand Limit ($L$.$H$.$L$.):$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{\sin(p+1)x + \sin x}{x} = \lim_{x 	o 0^-} \left( \frac{\sin(p+1)x}{x} + \frac{\sin x}{x} ight)$.Using $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,we get:$(p+1) + 1 = p+2$.Since $L.H.L. = q$,we have $p+2 = 1/2$,which implies $p = 1/2 - 2 = -3/2$.Therefore,the values are $(p, q) = (-3/2, 1/2)$.

The values of $p$ and $q$ for which the function $f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x} & x < 0 \\ q & x = 0 \\ \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} & x > 0 \end{cases}$ is continuous for $\forall x \in R$ are

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