mathop {Limit}limits_{x to 0^+} frac{1}{xsqrt | vedclass

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(D) Let $f(x) = a 	an^{-1} \frac{\sqrt{x}}{a} - b 	an^{-1} \frac{\sqrt{x}}{b}$ and $g(x) = x\sqrt{x} = x^{3/2}$.Since the limit is of the form $0/0$

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$\frac{a^2 - b^2}{3a^2b^2}$

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(D) Let $f(x) = a 	an^{-1} \frac{\sqrt{x}}{a} - b 	an^{-1} \frac{\sqrt{x}}{b}$ and $g(x) = x\sqrt{x} = x^{3/2}$.Since the limit is of the form $0/0$,we apply $L$'Hopital's rule:$\frac{d}{dx} (a 	an^{-1} \frac{\sqrt{x}}{a}) = a \cdot \frac{1}{1 + (\sqrt{x}/a)^2} \cdot \frac{1}{2a\sqrt{x}} = \frac{a^2}{a^2 + x} \cdot \frac{1}{2\sqrt{x}}$.Similarly,$\frac{d}{dx} (b 	an^{-1} \frac{\sqrt{x}}{b}) = \frac{b^2}{b^2 + x} \cdot \frac{1}{2\sqrt{x}}$.Derivative of $g(x) = \frac{3}{2}\sqrt{x}$.Applying the rule: $\lim_{x 	o 0^+} \frac{\frac{1}{2\sqrt{x}} (\frac{a^2}{a^2+x} - \frac{b^2}{b^2+x})}{\frac{3}{2}\sqrt{x}} = \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{a^2(b^2+x) - b^2(a^2+x)}{(a^2+x)(b^2+x)} ight)$.$= \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{a^2b^2 + a^2x - b^2a^2 - b^2x}{(a^2+x)(b^2+x)} ight) = \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{x(a^2 - b^2)}{(a^2+x)(b^2+x)} ight)$.$= \frac{a^2 - b^2}{3a^2b^2}$.

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

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