If the function f(x) = begin{cases} frac{x^2 | vedclass

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(B) For a function $f(x)$ to be continuous at $x = a$,the condition $\lim_{x 	o a} f(x) = f(a)$ must be satisfied.Here,$a = 1$,so we need $\lim_{x

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$2$

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(B) For a function $f(x)$ to be continuous at $x = a$,the condition $\lim_{x 	o a} f(x) = f(a)$ must be satisfied.Here,$a = 1$,so we need $\lim_{x 	o 1} f(x) = f(1)$.Given $f(1) = k$.Now,calculate the limit: $\lim_{x 	o 1} f(x) = \lim_{x 	o 1} \frac{x^2 - 1}{x - 1}$.Since this is a $0/0$ form,we factor the numerator: $\lim_{x 	o 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x 	o 1} (x + 1)$.Substituting $x = 1$,we get $1 + 1 = 2$.Therefore,for continuity,$k = 2$.

If the function $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \ne 1 \\ k, & x = 1 \end{cases}$ is continuous at $x = 1$,then the value of $k$ is:

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