Show that lim x->a (x^3/8-a^3/8)/(x^5/3-a^5/3)?

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Answer 1 · 2018-02-03T06:00:22

$lim _(x->a) (x^3/8-a^3/8)/(x^5/3-a^5/3)=(9) /(40a^(2))$

$lim _(x->a) (x^3/8-a^3/8)/(x^5/3-a^5/3)$

As we can easily recognize that this is $0/0$ we'll modify the fraction

$((x^3-a^3)*3)/((x^5-a^5 )*8)$

Apply the factoring rule

$(cancel(x -a)(a^2+ax+x^2)*3)/(8cancel(x-a)(x^4+x^3a+ x^2a^2+xa^3+a^4)$

Plug in the value a

$((a^2+aa+a^2)*3)/(8(a^4+a^3a+ a^2a^2+aa^3+a^4)$

$((3a^2)*3)/(8(2a^4+2a^3a^1+ a^2a^2)$

$(9a^2) /(8(2a^4+2a^4+ a^4)$

$(9a^2) /(8(5a^4)$

$(9a^2) /(40a^4)$

$= (9) /(40a^(4-2))$

$= (9) /(40a^(2))$

$lim _(x->a) (x^3/8-a^3/8)/(x^5/3-a^5/3)=(9) /(40a^(2))$