If f(x) = begin{cases} frac{x^4 - 16}{x - 2}, | vedclass

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(B) To check the continuity of $f(x)$ at $x = 2$,we evaluate the limit of $f(x)$ as $x 	o 2$: $\lim_{x 	o 2} f(x) = \lim_{x 	o 2} \frac{x^4 - 16}{x

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$f(x)$ is discontinuous at $x = 2$

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(B) To check the continuity of $f(x)$ at $x = 2$,we evaluate the limit of $f(x)$ as $x 	o 2$: $\lim_{x 	o 2} f(x) = \lim_{x 	o 2} \frac{x^4 - 16}{x - 2}$ Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$. Thus,$\lim_{x 	o 2} \frac{(x - 2)(x + 2)(x^2 + 4)}{x - 2} = \lim_{x 	o 2} (x + 2)(x^2 + 4) = (2 + 2)(2^2 + 4) = 4 	imes 8 = 32$. Since $\lim_{x 	o 2} f(x) = 32$ and $f(2) = 16$,we see that $\lim_{x 	o 2} f(x) 
eq f(2)$. Therefore,$f(x)$ is discontinuous at $x = 2$.

If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then:

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