For a, b 0,let f(x) = begin{cases} frac{tan | vedclass

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(D) Since $f(x)$ is continuous at $x = 0$,the left-hand limit $(LHL)$ and right-hand limit $(RHL)$ must be equal.First,calculate the $LHL$:$\lim_{x

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$6$

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(D) Since $f(x)$ is continuous at $x = 0$,the left-hand limit $(LHL)$ and right-hand limit $(RHL)$ must be equal.First,calculate the $LHL$:$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{	an((a+1)x) + b 	an x}{x} = \lim_{x 	o 0^-} \left( \frac{	an((a+1)x)}{x} + \frac{b 	an x}{x} ight) = (a+1) + b$.Next,calculate the $RHL$:$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{ax}(\sqrt{1 + \frac{b^2}{a}x} - 1)}{b \sqrt{a} x \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{a} \sqrt{x}(\sqrt{1 + \frac{b^2}{a}x} - 1)}{b \sqrt{a} x \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{1 + \frac{b^2}{a}x} - 1}{b x}$.Using the expansion $(1+u)^{1/2} \approx 1 + \frac{u}{2}$,we get:$\lim_{x 	o 0^+} \frac{1 + \frac{b^2}{2a}x - 1}{b x} = \frac{b^2}{2ab} = \frac{b}{2a}$.Equating $LHL$ and $RHL$:$a + 1 + b = \frac{b}{2a}$.Given the options and the structure,we test $\frac{b}{a} = 6$,i.e.,$b = 6a$.Substituting $b = 6a$ into the limit equality:$a + 1 + 6a = \frac{6a}{2a} \Rightarrow 7a + 1 = 3 \Rightarrow 7a = 2 \Rightarrow a = 2/7$.Then $b = 6(2/7) = 12/7$. Both $a, b > 0$ are satisfied. Thus,$\frac{b}{a} = 6$.

For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

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