<s>[INST] How to find constants a and b in this function? $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax+b}-2}x = \frac 5{12}$
How do you solve for constants $a$ and $b$?
 [/INST] 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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