find-the-values-of-x-for-which-x-3-8-x-2-4-is-discontinuous-and-state-each-kinds-of-dicontinuity-

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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