mathop {lim }limits_{x to 1} f(x) is equal to | vedclass

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(D) To find $\mathop {\lim }\limits_{x 	o 1} f(x)$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.$LHL = \mathop {\lim }\limits

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does not exist

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(D) To find $\mathop {\lim }\limits_{x 	o 1} f(x)$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.$LHL = \mathop {\lim }\limits_{x 	o 1^-} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2}$. Let $x = 1 - h$,where $h 	o 0^+$.$LHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{e^{-\frac{1}{h}} - 2}{e^{-\frac{1}{h}} + 2} = \frac{0 - 2}{0 + 2} = -1$.$RHL = \mathop {\lim }\limits_{x 	o 1^+} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2}$. Let $x = 1 + h$,where $h 	o 0^+$.$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{e^{\frac{1}{h}} - 2}{e^{\frac{1}{h}} + 2}$. Dividing numerator and denominator by $e^{\frac{1}{h}}$:$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{1 - 2e^{-\frac{1}{h}}}{1 + 2e^{-\frac{1}{h}}} = \frac{1 - 0}{1 + 0} = 1$.Since $LHL 
eq RHL$,the limit does not exist.

$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to,where $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

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