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Question Number 78693 by TawaTawa last updated on 19/Jan/20

Commented by abdomathmax last updated on 20/Jan/20

$l e t f (x) = \frac{(^{3} \sqrt{x}) + (^{3} \sqrt{a})}{x + a} c h a n g e m e n t x + a = t g i v e$ $f (x) = g (t) = \frac{(^{3} \sqrt{t - a} + (^{3} \sqrt{a})}{t} w e h s v e$ $α^{3} + β^{3} = (α + β) (α^{2} - α β + β^{2}) l e t c h a n g e α b y^{3} \sqrt{α}$ $a n d β b y^{3} \sqrt{β} w e g e t α + β = (^{3} \sqrt{α} +^{3} \sqrt{β}) (α^{\frac{2}{3}} -^{3} \sqrt{α β} + β^{\frac{2}{3}}) \Rightarrow$ $(^{3} \sqrt{α}) + (^{3} \sqrt{β}) = \frac{α + β}{α^{\frac{2}{3}} -^{3} \sqrt{α β} + β^{\frac{2}{3}}} \Rightarrow$ $(^{3} \sqrt{t - a}) + (^{3} \sqrt{a}) = \frac{t - a + a}{{(t - a)}^{\frac{2}{3}} -^{3} \sqrt{(t - a) a} + a^{\frac{2}{3}}} \Rightarrow$ $g (t) = \frac{1}{{(t - a)}^{\frac{2}{3}} -^{3} \sqrt{(t - a) a} + a^{\frac{2}{3}}} \Rightarrow$ ${l i m}_{t \to 0} g (t) = \frac{1}{{(- a)}^{\frac{2}{3}} + a^{\frac{2}{3}} + a^{\frac{2}{3}}} = \frac{1}{3 a^{\frac{2}{3}}} = \frac{1}{3 (^{3} \sqrt{a^{2})}}$ $= {l i m}_{x \to - a} f (x)$
Commented by TawaTawa last updated on 20/Jan/20

$God bless you sir$
Commented by mathmax by abdo last updated on 20/Jan/20

$y o u a r e w e l c o m e$
	Answered by john santu last updated on 20/Jan/20

	$w e k n o w t h a t$ $x + a = (\sqrt[3]{x} + \sqrt[3]{a}) (\sqrt[3]{x^{2}} - \sqrt[3]{a x} + \sqrt[3]{a})$ $\lim_{x \to - a} (\frac{1}{\sqrt[3]{x^{2}} - \sqrt[3]{a x} + \sqrt[3]{a^{2}}}) =$ $\frac{1}{\sqrt[3]{a^{2}} - \sqrt[3]{{(- a)}^{2}} + \sqrt[3]{a^{2}}} = \frac{1}{3 \sqrt[3]{a^{2}}} .$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir$
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