If-lim-x-3-17-ax-3-1-3-b-x-3-136-27-then-8a-b-

Question Number 115027 by bemath last updated on 23/Sep/20

If \lim_{x \to 3} \frac{17 \sqrt[3]{ax + 3} + b}{x - 3} = \frac{136}{27}

then 8 a + b = ?

Answered by bobhans last updated on 23/Sep/20

l i m i t f o r m \frac{0}{0} . n u m e r a t o r m u s t b e = 0

(1) 17 \sqrt[3]{3 a + 3} + b = 0; b \Rightarrow - 17 \sqrt[3]{3 a + 3}

(2) \lim_{x \to 3} \frac{17 \sqrt[3]{a x + 3} - 17 \sqrt[3]{3 a + 3}}{x - 3} = \frac{136}{27}

\lim_{x \to 3} \frac{\sqrt[3]{a x + 3} - \sqrt[3]{3 a + 3}}{x - 3} = \frac{8}{27}

s e t x = 3 + r; r \to 0

\lim_{r \to 0} \frac{\sqrt[3]{a r + 3 a + 3} - \sqrt[3]{3 a + 3}}{r} = \frac{8}{27}

r e m e m b e r \sqrt[3]{p} - \sqrt[3]{q} = \frac{p - q}{\sqrt[3]{p^{2}} + \sqrt[3]{p q} + \sqrt[3]{q^{2}}}

\lim_{r \to 0} \frac{(a r + 3 a + 3) - (3 a + 3)}{r (\sqrt[3]{{(a r + 3 a + 3)}^{2}} + \sqrt[3]{(3 a + 3) (a r + 3 a + 3)} + \sqrt[3]{{(3 a + 3)}^{2}}} = \frac{8}{27}

\lim_{r \to 0} \frac{a r}{r (\sqrt[3]{{(a r + 3 a + 3)}^{2}} + \sqrt[3]{(a r + 3 a + 3) (3 a + 3)} + \sqrt[3]{{(3 a + 3)}^{2}})} = \frac{8}{27}

\Leftrightarrow \frac{a}{3 \sqrt[3]{{(3 a + 3)}^{2}}} = \frac{8}{27};

b y i n s p e c t i o n w e g e t a = 8, c h e c k \frac{8}{3 {(\sqrt[3]{24 + 3})}^{2}}

= \frac{8}{3 \times 9} = \frac{8}{27} . T h e n b = - 17 \sqrt[3]{24 + 3} = - 51

∴ 8 a + b = 64 - 51 = 13

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