If f(x) = begin{cases} frac{tan(2p-7)x + tan | vedclass

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(A) Since the function is continuous at $x=0$, we have $f(0^+) = f(0^-) = f(0)$.First, calculate the right-hand limit $f(0^+)$:$f(0^+) = \lim_{x 	o 0

Vedclass · Accepted Answer

$\frac{2}{3}$

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(A) Since the function is continuous at $x=0$, we have $f(0^+) = f(0^-) = f(0)$.First, calculate the right-hand limit $f(0^+)$:$f(0^+) = \lim_{x 	o 0^+} q \left( \frac{\sqrt{x^2+x} - \sqrt{x}}{x^{3/2}} ight) = q \lim_{x 	o 0^+} \frac{(\sqrt{x^2+x} - \sqrt{x})(\sqrt{x^2+x} + \sqrt{x})}{x^{3/2}(\sqrt{x^2+x} + \sqrt{x})}$$= q \lim_{x 	o 0^+} \frac{x^2+x-x}{x^{3/2}(\sqrt{x}(\sqrt{x+1}+1))} = q \lim_{x 	o 0^+} \frac{x^2}{x^2(\sqrt{x+1}+1)} = \frac{q}{2}$.Next, calculate the left-hand limit $f(0^-)$:$f(0^-) = \lim_{x 	o 0^-} \frac{	an(2p-7)x + 	an 3x}{x} = \lim_{x 	o 0^-} \left( \frac{	an(2p-7)x}{x} + \frac{	an 3x}{x} ight) = (2p-7) + 3 = 2p-4$.Given $f(0) = p-q$, for continuity at $x=0$:$f(0^-) = f(0) \implies 2p-4 = p-q \implies p+q = 4$.$f(0^+) = f(0) \implies \frac{q}{2} = p-q \implies q = 2p-2q \implies 3q = 2p$.From $3q = 2p$, we get $\frac{q}{p} = \frac{2}{3}$.

If $f(x) = \begin{cases} \frac{\tan(2p-7)x + \tan 3x}{x}, & x < 0 \\ p-q, & x=0 \\ q\left(\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}\right), & x > 0 \end{cases}$. If $f(x)$ is continuous at $x=0$, then $\frac{q}{p} = $

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