If the function f(x) satisfies mathop {lim }l | vedclass

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(C) Given that $\mathop {\lim }\limits_{x 	o 1} \frac{f(x)-2}{x^{2}-1}=\pi.$For the limit to exist and be a finite value $\pi$,the numerator must app

Vedclass · Accepted Answer

$2$

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(C) Given that $\mathop {\lim }\limits_{x 	o 1} \frac{f(x)-2}{x^{2}-1}=\pi.$For the limit to exist and be a finite value $\pi$,the numerator must approach $0$ as $x 	o 1$ because the denominator $x^2 - 1$ approaches $0$.Therefore,$\mathop {\lim }\limits_{x 	o 1} (f(x) - 2) = 0.$This implies $\mathop {\lim }\limits_{x 	o 1} f(x) - \mathop {\lim }\limits_{x 	o 1} 2 = 0.$Since $\mathop {\lim }\limits_{x 	o 1} 2 = 2$,we have $\mathop {\lim }\limits_{x 	o 1} f(x) - 2 = 0.$Thus,$\mathop {\lim }\limits_{x 	o 1} f(x) = 2.$

If the function $f(x)$ satisfies $\mathop {\lim }\limits_{x \to 1} \frac{f(x)-2}{x^{2}-1}=\pi,$ evaluate $\mathop {\lim }\limits_{x \to 1} f(x).$

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