find the values of x for which ((x^3 +8)/(x^2 −4 ))is discontinuous and state each kinds of dicontinuity


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Question Number 12651 by okhemafrancis last updated on 28/Apr/17

$f i n d t h e v a l u e s o f x f o r w h i c h \frac{x^{3} + 8}{x^{2} - 4} i s d i s c o n t i n u o u s a n d s t a t e e a c h k i n d s o f d i c o n t i n u i t y$
	Answered by FilupS last updated on 28/Apr/17

	$f (x) = \frac{x^{3} + 8}{x^{2} - 4}$ $for f : x \to R, - \infty ⩽ x ⩽ x \Leftrightarrow x^{2} - 4 \neq 0$ $∴ x^{2} \neq 4$ $x \neq \pm 2$ $for x = - 2 \Rightarrow \frac{x^{3} + 8}{x^{2} - 4} = \frac{- 8 + 8}{4 - 4} = \frac{0}{0}$ $L^{'} Hospital Rule$ $\lim_{x \to - 2} \frac{x^{3} + 8}{x^{2} - 4} = \lim_{x \to - 2} \frac{3 x^{2}}{2 x} = \frac{2 {(- 2)}^{2}}{2} = 4$ $∴ x = - 2 is a solution$ $i . e . x \neq + 2$ $for x = + 2$ $\lim_{x \to 2^{-}} f = - \infty and \lim_{x \to 2^{+}} f = + \infty$
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