mathop {lim }limits_{x to 0} sin left( {frac{ | vedclass

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(D) Let $f(x) = \sin \left( \frac{1}{x} \right)$.For the limit to exist at $x = 0$,the left-hand limit and right-hand limit must be equal.Left-hand li

Vedclass · Accepted Answer

Does not exist

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(D) Let $f(x) = \sin \left( \frac{1}{x} ight)$.For the limit to exist at $x = 0$,the left-hand limit and right-hand limit must be equal.Left-hand limit: $\mathop {\lim }\limits_{x 	o {0^ - }} f(x) = \mathop {\lim }\limits_{h 	o 0} \sin \left( \frac{1}{-h} ight) = \mathop {\lim }\limits_{h 	o 0} -\sin \left( \frac{1}{h} ight)$.As $h 	o 0$,$\frac{1}{h} 	o \infty$,and $\sin \left( \frac{1}{h} ight)$ oscillates between $-1$ and $1$. Thus,the limit does not exist.Right-hand limit: $\mathop {\lim }\limits_{x 	o {0^ + }} f(x) = \mathop {\lim }\limits_{h 	o 0} \sin \left( \frac{1}{h} ight)$.Similarly,as $h 	o 0$,$\sin \left( \frac{1}{h} ight)$ oscillates between $-1$ and $1$.Since the left-hand limit and right-hand limit do not approach a unique finite value,$\mathop {\lim }\limits_{x 	o 0} \sin \left( \frac{1}{x} ight)$ does not exist.

$\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ is

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