If f(x) = begin{cases} x^2, & text{when } x l | vedclass

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(B) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.$LHL$: $\lim_{x 	o 1^-} f(x

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

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(B) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.$LHL$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (x^2) = (1)^2 = 1$.$RHL$: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (x + 5) = 1 + 5 = 6$.Since the left-hand limit $(1)$ is not equal to the right-hand limit $(6)$,the limit $\lim_{x 	o 1} f(x)$ does not exist.Therefore,$f(x)$ is discontinuous at $x = 1$.

If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

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