Let [x] denote the greatest integer less than | vedclass

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(A) We evaluate the limit $L = \mathop {\lim }\limits_{x 	o 0} \frac{{	an (\pi {{\sin }^2}x) + (\left| x ight| - \sin (x[x]))^2}}{{{x^2}}}$.First,

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does not exist

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(A) We evaluate the limit $L = \mathop {\lim }\limits_{x 	o 0} \frac{{	an (\pi {{\sin }^2}x) + (\left| x ight| - \sin (x[x]))^2}}{{{x^2}}}$.First,consider the term $\mathop {\lim }\limits_{x 	o 0} \frac{{	an (\pi {{\sin }^2}x)}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{	an (\pi {{\sin }^2}x)}}{{\pi {{\sin }^2}x}} \cdot \frac{{\pi {{\sin }^2}x}}{{{x^2}}} ight) = 1 \cdot \pi \cdot 1^2 = \pi$.Now consider the second term: $\mathop {\lim }\limits_{x 	o 0} \frac{{(\left| x ight| - \sin (x[x]))^2}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\left| x ight| - \sin (x[x])}}{{\left| x ight|}} ight)^2$ (since $x^2 = |x|^2$).For $x 	o 0^+$,$[x] = 0$,so $\mathop {\lim }\limits_{x 	o 0^+} \left( 1 - \frac{{\sin (x \cdot 0)}}{x} ight)^2 = (1 - 0)^2 = 1$.For $x 	o 0^-$,$[x] = -1$,so $\mathop {\lim }\limits_{x 	o 0^-} \left( 1 - \frac{{\sin (-x)}}{{-x}} ight)^2 = (1 - 1)^2 = 0$.Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan \,(\pi \,{{\sin }^2}\,x) + \,{{(\left| x \right|\, - \,\sin \,(x\,[x]))}^2}}}{{{x^2}}}$

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