91Ó°ÊÓ

First, let's plug in \(x=0\) into the expression to check if it's an indeterminate form (which means, it's either of the form 0/0, ∞/∞, or similar forms): \(\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\tan ^{-1} 3 x}-\sqrt[3]{1-\sin^{-1} 3 x}}{\sqrt{1-\sin^{-1} 2 x}-\sqrt{1+\tan^{-1} 2 x}} = \frac{\sqrt[3]{1}-\sqrt[3]{1}}{\sqrt{1}-\sqrt{1}}\) Since we have a 0/0 form, we can use L'Hopital's Rule. First, differentiate the numerator and the denominator with respect to x. Differentiate the numerator: $u(x) = \sqrt[3]{1+\tan ^{-1} 3 x}\newline v(x) = \sqrt[3]{1-\sin^{-1} 3 x}$ $u'(x) = \frac{1}{\sqrt[3]{(1+\tan ^{-1} 3 x)^2}} * \frac{1}{1+(3x)^2}*3\newline v'(x) = -\frac{1}{\sqrt[3]{(1-\sin^{-1} 3 x)^2}} * \frac{1}{\sqrt{1-(3x)^2}}*3$ Differentiate the denominator: $m(x) = \sqrt{1-\sin ^{-1} 2 x}\newline n(x) = \sqrt{1+\tan ^{-1} 2 x}$ $m'(x) = -\frac{1}{2\sqrt{1-\sin^{-1} 2 x}} * \frac{1}{\sqrt{1-(2x)^2}}*2\newline n'(x) = \frac{1}{2\sqrt{1+\tan^{-1} 2 x}} * \frac{1}{1+(2x)^2}*2$ Now, we will apply L'Hopital's rule: \(\lim _{x \rightarrow 0} \frac{u'(x) - v'(x)}{m'(x)-n'(x)}\).

To simplify the expression, we will use the following trigonometric limits: 1. \(\lim_{x \rightarrow 0} \frac{\tan^{-1}(ax)}{x} = a\) 2. \(\lim_{x \rightarrow 0} \frac{\sin^{-1}(ax)}{x} = a\) Applying above limits, we get: \(\lim_{x \rightarrow 0} \frac{\frac{3}{\sqrt[3]{(1+\tan ^{-1} 3 x)^2}} * \frac{1}{1+(3x)^2}-[-\frac{3}{\sqrt[3]{(1-\sin^{-1} 3 x)^2}} * \frac{1}{\sqrt{1-(3x)^2}}]}{-\frac{1}{\sqrt{1-\sin^{-1} 2 x}} * \frac{1}{\sqrt{1-(2x)^2}}+\frac{1}{\sqrt{1+\tan^{-1} 2 x}} * \frac{1}{1+(2x)^2}}\)

Now, we will cancel out the common factors and evaluate the limit: \(\lim_{x \rightarrow 0} \frac{3\sqrt[3]{(1-\sin^{-1} 3 x)^2} * (1+(3x)^2) + 3\sqrt[3]{(1+\tan ^{-1} 3 x)^2} * \sqrt{1-(3x)^2}}{1+(2x)^2}\) Plug \(x=0\) in the expression: $\frac{3\sqrt[3]{1^2} * (1+0) + 3\sqrt[3]{1^2} * \sqrt{1}}{1+0} =\frac{3+3}{1} =6$ So none of the given options (A, B, and C) match the result. Therefore, our answer is: (D) None