Question

\(\lim\limits_{x\rightarrow0}\frac{\sqrt{8x^3+x^2+6x+9}-\sqrt[3]{9x^2+27x+27}}{x^3}\)

Answer 1

\(\lim\limits_{x\rightarrow0}\frac{\left(\sqrt{8x^3+x^2+6x+9}-\left(x+3\right)\right)+\left(x+3-\sqrt[3]{9x^2+27x+27}\right)}{x^3}\)

\(=\lim\limits_{x\rightarrow0}\frac{\frac{8x^3}{\sqrt{8x^3+x^2+6x+9}+x+3}+\frac{x^3}{\left(x+3\right)^2+\left(x+3\sqrt[3]{9x^2+27x+27}+\sqrt[3]{\left(9x^2+27x+27\right)^2}\right)}}{x^3}\)

\(=\lim\limits_{x\rightarrow0}\left(\frac{8}{\sqrt{8x^3+x^2+6x+9}+x+3}+\frac{1}{\left(x+3\right)^2+\left(x+3\sqrt[3]{9x^2+27x+27}+\sqrt[3]{\left(9x^2+27x+27\right)^2}\right)}\right)\)

\(=\frac{8}{3+3}+\frac{1}{9+3.3+\sqrt[3]{27^2}}=\frac{37}{27}\)