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$\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax+b}-2}x = \frac 5{12}$

How do you solve for constants $a$ and $b$?

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As the denominator goes $\to 0$, so must the numerator, hence $\sqrt[3]{ax+b}\to 2$ and $ax+b\to 8$ This gives you $b=8$. Note that this allows you to interpete the limit as $f'(0)$ where $f'(x)=\sqrt[3]{ax+8}$, whence you can find $a$.

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You can use $(x^3-y^3)=(x-y)(x^2+xy+y^2)$ with $x=\sqrt[3]{ax+b},y=2$ and multiply the term $\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)$ into nominator and denominator of the limit: $$\displaystyle \lim_{x\to0} \frac{\big(\sqrt[3]{ax+b}-2\big)\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)} = \lim_{x\to0} \frac {ax+b-8}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}\\= \lim_{x\to0} \frac {x(a+\frac{b-8}{x})}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}=\lim_{x\to0} \frac {(a+\frac{b-8}{x})}{\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}$$ which should be definite and equals to $5/12$. So the term $(a+\frac{b-8}{x})$ should be certain and then $b$ must be $8$ at $x=0$. Now put $b=8$ in your fraction: $$\lim_{x\to0} \frac {a}{\big(\sqrt[3]{(ax+8)^2}+2\sqrt[3]{ax+8}+4\big)}$$ which is $5/12$. I think it is easy to find $a$.

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Nicely done! $\quad\ddot\smile \quad +1\;\;$ –  amWhy Mar 30 '13 at 1:01
    
@amWhy: Thanks Amy. I hope you have had a good day, up yo now. ;-) –  B. S. Mar 30 '13 at 1:53
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