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

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

$${find}\:{the}\:{values}\:{of}\:{x}\:{for}\:{which}\:\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}\:}{is}\:{discontinuous}\:{and}\:{state}\:{each}\:{kinds}\:{of}\:{dicontinuity} \\ $$

Answered by FilupS last updated on 28/Apr/17

f(x)=((x^3 +8)/(x^2 −4)) for f:x→R, −∞≤x≤x ⇔ x^2 −4≠0 ∴x^2 ≠4 x≠±2 for x=−2 ⇒ ((x^3 +8)/(x^2 −4)) = ((−8+8)/(4−4))=(0/0) L′Hospital Rule lim_(x→−2) ((x^3 +8)/(x^2 −4))=lim_(x→−2) ((3x^2 )/(2x))=((2(−2)^2 )/2)=4 ∴x=−2 is a solution i.e. x≠+2 for x=+2 lim_(x→2^− ) f=−∞ and lim_(x→2^+ ) f=+∞

$${f}\left({x}\right)=\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$$$\mathrm{for}\:{f}:{x}\rightarrow\mathbb{R},\:\:\:−\infty\leqslant{x}\leqslant{x}\:\:\:\Leftrightarrow\:\:\:{x}^{\mathrm{2}} −\mathrm{4}\neq\mathrm{0} \\ $$$$\therefore{x}^{\mathrm{2}} \neq\mathrm{4} \\ $$$${x}\neq\pm\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{for}\:\:{x}=−\mathrm{2}\:\:\Rightarrow\:\:\:\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}}\:\:=\:\frac{−\mathrm{8}+\mathrm{8}}{\mathrm{4}−\mathrm{4}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\mathrm{L}'\mathrm{Hospital}\:\mathrm{Rule} \\ $$$$\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}}=\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\frac{\mathrm{2}\left(−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{4} \\ $$$$\: \\ $$$$\therefore{x}=−\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{i}.\mathrm{e}.\:\:{x}\neq+\mathrm{2} \\ $$$$\: \\ $$$$\: \\ $$$$\mathrm{for}\:{x}=+\mathrm{2} \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}{f}=−\infty\:\:\:\:\mathrm{and}\:\:\:\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}{f}=+\infty \\ $$

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