Question

plz answer fast
answer should be 2/3.​

Answer 1

Step-by-step explanation:

$\huge\blue{\underline{\underline{\bf Question}}}$

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$\displaystyle \lim_{x \: \to \: 0} \frac{ {(1 + x)}^{ \frac{1}{3} } - {(1 - x)}^{ \frac{1}{3} } }{x}$

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$\huge\blue{\underline{\underline{\bf Answer}}}$

$\implies \displaystyle \lim_{x \: \to \: 0} \frac{ {(1 + x)}^{ \frac{1}{3} } - {(1 - x)}^{ \frac{1}{3} } }{x} \\ \\ we \: know \: that = > \\ \\ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ \\ let \: a = {(1 + x)}^{ \frac{1}{3} } \: \: and \: \: b = {(1 - x)}^{ \frac{1}{3} } \\ \\ \implies {({(1 + x)}^{ \frac{1}{3} })}^{3} - {({(1 - x)}^{ \frac{1}{3} })}^{3} = \\ {(1 + x)}^{ \frac{1}{3} } - {(1 - x)}^{ \frac{1}{3} } ( {(1 + x)}^{ \frac{2}{3} } + {(1 - x)}^{ \frac{2}{3} } {(1 - x)}^{ \frac{1}{3} } \\ \\ \implies \displaystyle \lim_{x \to \: 0} \frac{{(1 + x)}^{ \frac{1}{3} } - {(1 - x)}^{ \frac{1}{3} }}{x} = \frac{ {({(1 + x)}^{ \frac{1}{3} })}^{3} - {({(1 - x)}^{ \frac{1}{3} })}^{3} }{ x ( {(1 + x)}^{ \frac{2}{3} } + {(1 - x)}^{ \frac{2}{3} } {(1 - x)}^{ \frac{1}{3} } ) } \\ \\ \implies \frac{2 \cancel{x}}{ \cancel{x} {(1)}^{ \frac{2}{3} } + {(1)}^{ \frac{2}{3} } + {(1)}^{ \frac{1}{3} } } \\ \\ \implies \frac{2}{1 + 1 + 1} \\ \\ \implies \huge \boxed{ \frac{2}{3} }$