Let f(x) = begin{cases} frac{tan^2 {x}}{x^2 - | vedclass

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(D) For $x 	o 0^+$,$[x] = 0$ and $\{x\} = x$. Thus,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{	an^2 x}{x^2 - 0^2} = \lim_{x 	o 0^+} \left( \fr

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Both $(A)$ and $(C)$

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(D) For $x 	o 0^+$,$[x] = 0$ and $\{x\} = x$. Thus,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{	an^2 x}{x^2 - 0^2} = \lim_{x 	o 0^+} \left( \frac{	an x}{x} ight)^2 = 1^2 = 1$.For $x 	o 0^-$,let $x = -h$ where $h > 0$. Then $[x] = -1$ and $\{x\} = x - [x] = -h - (-1) = 1 - h$.$\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0^+} \sqrt{(1-h) \cot(1-h)} = \sqrt{1 \cdot \cot(1)} = \sqrt{\cot 1}$.Now,$\left( \lim_{x 	o 0^-} f(x) ight)^2 = (\sqrt{\cot 1})^2 = \cot 1$.Therefore,$\cot^{-1} \left( \lim_{x 	o 0^-} f(x) ight)^2 = \cot^{-1}(\cot 1) = 1$.Thus,both $(A)$ and $(C)$ are correct.

Let $f(x) = \begin{cases} \frac{\tan^2 \{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot \{x\}} & \text{for } x < 0 \end{cases}$ where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

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