If f(x) = begin{cases} frac{x^2 - 1}{x + 1}, | vedclass

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(D) Given $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x 
eq -1 \ -2, & x = -1 \end{cases}$.For $x 
eq -1$,$f(x) = \frac{(x - 1)(x + 1)}{x + 1} =

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All the above are correct

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(D) Given $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x 
eq -1 \ -2, & x = -1 \end{cases}$.For $x 
eq -1$,$f(x) = \frac{(x - 1)(x + 1)}{x + 1} = x - 1$.Now,calculate the left-hand limit: $\lim_{x 	o (-1)^-} f(x) = \lim_{x 	o -1} (x - 1) = -1 - 1 = -2$.Calculate the right-hand limit: $\lim_{x 	o (-1)^+} f(x) = \lim_{x 	o -1} (x - 1) = -1 - 1 = -2$.Since $\lim_{x 	o (-1)^-} f(x) = \lim_{x 	o (-1)^+} f(x) = f(-1) = -2$,the function is continuous at $x = -1$.Therefore,all the given statements are correct.

If $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x \neq -1 \\ -2, & x = -1 \end{cases}$,then which of the following is true?

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