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

$$\mathrm{If}\underset{x\to 3}{\lim }\frac{17\sqrt[3]{\mathrm{ax}+3}+\mathrm{b}}{\mathrm{x}-3}=\frac{136}{27}$$$$\mathrm{then}8\mathrm{a}+\mathrm{b}=?$$

Answered by bobhans last updated on 23/Sep/20

$$limitform\frac{0}{0}.numeratormustbe=0$$$$\left(1\right)17\sqrt[3]{3a+3}+b=0;b\Rightarrow -17\sqrt[3]{3a+3}$$$$\left(2\right)\underset{x\to 3}{\lim }\frac{17\sqrt[3]{ax+3}-17\sqrt[3]{3a+3}}{x-3}=\frac{136}{27}$$$$\underset{x\to 3}{\lim }\frac{\sqrt[3]{ax+3}-\sqrt[3]{3a+3}}{x-3}=\frac{8}{27}$$$$setx=3+r;r\to 0$$$$\underset{r\to 0}{\lim }\frac{\sqrt[3]{ar+3a+3}-\sqrt[3]{3a+3}}{r}=\frac{8}{27}$$$$remember\sqrt[3]{p}-\sqrt[3]{q}=\frac{p-q}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}$$$$\underset{r\to 0}{\lim }\frac{(ar+3a+3)-(3a+3)}{r(\sqrt[3]{{(ar+3a+3)}^2}+\sqrt[3]{(3a+3)(ar+3a+3)}+\sqrt[3]{{(3a+3)}^2}}=\frac{8}{27}$$$$\underset{r\to 0}{\lim }\frac{ar}{r(\sqrt[3]{{(ar+3a+3)}^2}+\sqrt[3]{(ar+3a+3)(3a+3)}+\sqrt[3]{{(3a+3)}^2})}=\frac{8}{27}$$$$\iff \frac{a}{3\sqrt[3]{{(3a+3)}^2}}=\frac{8}{27};$$$$byinspectionwegeta=8,check\frac{8}{3{\left(\sqrt[3]{24+3}\right)}^2}$$$$=\frac{8}{3\times 9}=\frac{8}{27}.Thenb=-17\sqrt[3]{24+3}=-51$$$$\therefore 8a+b=64-51=13$$

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