4. $3 x=\left(x-\frac{2}{3}\right)+(x+1)+\left(x-\frac{1}{3}\right)$, from which

$$
\begin{aligned}
& L=\lim _{x \rightarrow \infty} x\left(\sqrt[3]{x^{3}-2 x^{2}+a x+b}-\left(x-\frac{2}{3}\right)\right)+\lim _{x \rightarrow \infty} x\left(\sqrt[3]{x^{3}+3 x^{2}+\alpha x+\beta}-(x+1)\right)+ \\
& +\lim _{x \rightarrow \infty} x\left(\sqrt[3]{x^{3}-x^{2}+m x+n}-\left(x-\frac{1}{3}\right)\right)=L_{1}+L_{2}+L_{3}
\end{aligned}
$$

By amplifying with the conjugate expression, each term will reduce the $x^{3}$ and $x^{2}$ terms and we obtain

$$
\begin{aligned}
& L_{1}=\lim _{x \rightarrow \infty} x \cdot \frac{x^{3}-2 x^{2}+a x+b-x^{3}+2 x^{2}-\frac{4}{3} x+\frac{8}{27}}{\left(\sqrt[3]{x^{3}-2 x^{2}+a x+b}\right)^{2}+\left(x-\frac{2}{3}\right) \sqrt[3]{x^{3}-2 x^{2}+a x+b}+\left(x-\frac{2}{3}\right)^{2}}= \\
& =\frac{\left(a-\frac{4}{3}\right) x^{2}+\left(b+\frac{8}{27}\right) x}{\ldots}=\frac{a-\frac{4}{3}}{3}
\end{aligned}
$$

Similarly

$L_{2}=\lim _{x \rightarrow \infty} \frac{(\alpha-3) x^{2}+(\beta-1) x}{\ldots}=\frac{\alpha-3}{3}$ and $L_{3}=\lim _{x \rightarrow \infty} \frac{\left(m-\frac{1}{3}\right) x^{2}+\left(n+\frac{1}{27}\right) x}{\ldots}=\frac{m-\frac{1}{3}}{3}$, from which

$L=\frac{3 a+3 \alpha+3 m-14}{9}$

(3p)

## SIBIU COUNTY SCHOOL INSPECTORATE

## MATHEMATICS OLYMPIAD  LOCAL PHASE, 9.02.2013

Grade 11
4. (7p) Calculate $\lim _{x \rightarrow \infty} x\left(\sqrt[3]{x^{3}-2 x^{2}+a x+b}+\sqrt[3]{x^{3}+3 x^{2}+\alpha x+\beta}+\sqrt[3]{x^{3}-x^{2}+m x+n}-3 x\right)$, where $a$, $b, \alpha, \beta, m, n$ are real numbers.

Livia Băcilă

Note: All problems are mandatory.

Effective working time: 3 hours.