mathop {lim }limits_{x to 2} frac{|x - 2|}{x | vedclass

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(C) To find the limit $\mathop {\lim }\limits_{x 	o 2} \frac{|x - 2|}{x - 2}$,we evaluate the left-hand limit and the right-hand limit.Left-hand limi

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(C) To find the limit $\mathop {\lim }\limits_{x 	o 2} \frac{|x - 2|}{x - 2}$,we evaluate the left-hand limit and the right-hand limit.Left-hand limit $(LHL)$: $\mathop {\lim }\limits_{x 	o 2^-} \frac{|x - 2|}{x - 2} = \mathop {\lim }\limits_{h 	o 0} \frac{|2 - h - 2|}{2 - h - 2} = \mathop {\lim }\limits_{h 	o 0} \frac{|-h|}{-h} = \mathop {\lim }\limits_{h 	o 0} \frac{h}{-h} = -1$.Right-hand limit $(RHL)$: $\mathop {\lim }\limits_{x 	o 2^+} \frac{|x - 2|}{x - 2} = \mathop {\lim }\limits_{h 	o 0} \frac{|2 + h - 2|}{2 + h - 2} = \mathop {\lim }\limits_{h 	o 0} \frac{|h|}{h} = \mathop {\lim }\limits_{h 	o 0} \frac{h}{h} = 1$.Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

$\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2} = $

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