If f(x) = begin{cases} frac{x^2 - 4x + 3}{x^2 | vedclass

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(C) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 1$.Given $f(1)

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$f(x)$ is discontinuous at $x = 1$

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(C) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 1$.Given $f(1) = 2$.For $x 
e 1$,$f(x) = \frac{x^2 - 4x + 3}{x^2 - 1} = \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} = \frac{x - 3}{x + 1}$.Now,calculate the limit as $x 	o 1$:$\lim_{x 	o 1} f(x) = \lim_{x 	o 1} \frac{x - 3}{x + 1} = \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1$.Since $\lim_{x 	o 1^+} f(x) = -1$ and $\lim_{x 	o 1^-} f(x) = -1$,the limit exists and is equal to $-1$.However,$f(1) = 2$.Since $\lim_{x 	o 1} f(x) 
e f(1)$,the function $f(x)$ is discontinuous at $x = 1$.

If $f(x) = \begin{cases} \frac{x^2 - 4x + 3}{x^2 - 1}, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then:

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