mathop {Lim}limits_{x to c} f(x) does not exi | vedclass

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(D) Option $A$: $f(x) = [[x]] - [2x - 1]$ at $c = 3$. Right hand limit: $\lim_{h 	o 0} f(3+h) = \lim_{h 	o 0} [[3+h]] - [2(3+h)-1] = 3 - 5 = -2$. Le

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Both $(B)$ and $(C)$

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(D) Option $A$: $f(x) = [[x]] - [2x - 1]$ at $c = 3$. Right hand limit: $\lim_{h 	o 0} f(3+h) = \lim_{h 	o 0} [[3+h]] - [2(3+h)-1] = 3 - 5 = -2$. Left hand limit: $\lim_{h 	o 0} f(3-h) = \lim_{h 	o 0} [[3-h]] - [2(3-h)-1] = 2 - 4 = -2$. Since $LHL = RHL$,the limit exists. Option $B$: $f(x) = [x] - x$ at $c = 1$. Right hand limit: $\lim_{h 	o 0} f(1+h) = \lim_{h 	o 0} [1+h] - (1+h) = 1 - 1 = 0$. Left hand limit: $\lim_{h 	o 0} f(1-h) = \lim_{h 	o 0} [1-h] - (1-h) = 0 - 1 = -1$. Since $LHL 
eq RHL$,the limit does not exist. Option $C$: $f(x) = \{x\}^2 - \{-x\}^2$ at $c = 0$. Right hand limit: $\lim_{h 	o 0} f(0+h) = \lim_{h 	o 0} (h)^2 - (1-h)^2 = 0 - 1 = -1$. Left hand limit: $\lim_{h 	o 0} f(0-h) = \lim_{h 	o 0} (1-h)^2 - (h)^2 = 1 - 0 = 1$. Since $LHL 
eq RHL$,the limit does not exist. Therefore,both $(B)$ and $(C)$ are correct.

$\mathop {Lim}\limits_{x \to c} f(x)$ does not exist when: (where $[x]$ denotes the greatest integer function and $\{x\}$ denotes the fractional part function.)

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