The value of mathop {text{Limit}}limits_{x to | vedclass

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(D) Let $f(x) = \frac{	an(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$.For the right-hand limit $(x 	o 0^+)$,$\{x\} = x$ where $x$ is a small positive v

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non-existent

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(D) Let $f(x) = \frac{	an(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$.For the right-hand limit $(x 	o 0^+)$,$\{x\} = x$ where $x$ is a small positive value $h$:$\mathop {	ext{Limit}}\limits_{h 	o 0^+} \frac{	an(h - 1) \sin h}{h(h - 1)} = \frac{	an(-1)}{-1} 	imes \mathop {	ext{Limit}}\limits_{h 	o 0^+} \frac{\sin h}{h} = 	an 1 	imes 1 = 	an 1$.For the left-hand limit $(x 	o 0^-)$,$\{x\} = 1 + x$ where $x$ is a small negative value $-h$ $(h > 0)$:$\mathop {	ext{Limit}}\limits_{h 	o 0^+} \frac{	an(1 - h - 1) \sin(1 - h)}{(1 - h)(1 - h - 1)} = \mathop {	ext{Limit}}\limits_{h 	o 0^+} \frac{	an(-h) \sin(1 - h)}{(1 - h)(-h)} = \frac{\sin 1}{1} = \sin 1$.Since the right-hand limit $(	an 1)$ is not equal to the left-hand limit $(\sin 1)$,the limit does not exist.

The value of $\mathop {\text{Limit}}\limits_{x \to 0} \frac{\tan(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$,where $\{x\}$ denotes the fractional part function,is:

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