If f(x) = begin{cases} 2^{1/x}, & x ne 0 3, | vedclass

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(D) To determine the behavior of the function at $x = 0$,we calculate the left-hand limit and the right-hand limit.Right-hand limit $(RHL)$:$\mathop {

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None of these

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(D) To determine the behavior of the function at $x = 0$,we calculate the left-hand limit and the right-hand limit.Right-hand limit $(RHL)$:$\mathop {\lim }\limits_{x 	o 0^+} f(x) = \mathop {\lim }\limits_{h 	o 0} 2^{1/(0+h)} = \mathop {\lim }\limits_{h 	o 0} 2^{1/h} = \infty$.Left-hand limit $(LHL)$:$\mathop {\lim }\limits_{x 	o 0^-} f(x) = \mathop {\lim }\limits_{h 	o 0} 2^{1/(0-h)} = \mathop {\lim }\limits_{h 	o 0} 2^{-1/h} = \mathop {\lim }\limits_{h 	o 0} \frac{1}{2^{1/h}} = \frac{1}{\infty} = 0$.Since the $RHL$ is $\infty$ and the $LHL$ is $0$,the limit $\mathop {\lim }\limits_{x 	o 0} f(x)$ does not exist.Therefore,none of the given options $(A)$,$(B)$,or $(C)$ are correct.

If $f(x) = \begin{cases} 2^{1/x}, & x \ne 0 \\ 3, & x = 0 \end{cases}$,then:

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