The function f(x) = frac{x^2 - 1}{x^3 - 1} is | vedclass

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(B) For the function $f(x)$ to be continuous at $x = 1$,the value $f(1)$ must be equal to the limit of the function as $x$ approaches $1$.We have $f(x

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$\frac{2}{3}$

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(B) For the function $f(x)$ to be continuous at $x = 1$,the value $f(1)$ must be equal to the limit of the function as $x$ approaches $1$.We have $f(x) = \frac{x^2 - 1}{x^3 - 1}$.Factorizing the numerator and the denominator:$f(x) = \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)}$.For $x 
eq 1$,we can cancel the common factor $(x - 1)$:$f(x) = \frac{x + 1}{x^2 + x + 1}$.Now,taking the limit as $x 	o 1$:$\lim_{x 	o 1} f(x) = \lim_{x 	o 1} \frac{x + 1}{x^2 + x + 1} = \frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{3}$.Therefore,for the function to be continuous at $x = 1$,$f(1) = \frac{2}{3}$.

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

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