Fun with 2018 #8

$\boxed{1}$ Let $a$ be areal number, $\displaystyle \lim_{x \to \infty} \left ( \sqrt[3]{x^3 + ax^2} - \sqrt[3]{x^3 - ax^2} \right ) = \lim_{x \to \infty} \frac{\left ( \sqrt[3]{x^3 + ax^2} - \sqrt[3]{x^3 - ax^2} \right ) \cdot \left ( \sqrt[3]{(x^3 + ax^2)^2} + \sqrt[3]{(x^3 + ax^2)} \sqrt{(x^3 -ax^2)} + \sqrt[3]{(x^3 - ax^2)^2} \right )}{\left ( \sqrt[3]{(x^3 + ax^2)^2} + \sqrt[3]{(x^3 + ax^2)} \sqrt{(x^3 -ax^2)} + \sqrt[3]{(x^3 - ax^2)^2} \right )} =$ $= \displaystyle \lim_{x \to \infty} \frac{\left ( (x^3 + ax^2) - (x^3 - ax^2) \right )}{\left ( \sqrt[3]{(x^3 + ax^2)^2} + \sqrt[3]{(x^3 + ax^2)} \sqrt{(x^3 -ax^2)} + \sqrt[3]{(x^3 - ax^2)^2} \right )} =$ $= \displaystyle \lim_{x \to \infty} \frac{2ax^2}{\left ( \sqrt[3]{(x^6 + 2ax^5 + a^2 x^4)} + \sqrt[3]{(x^6 - a^2 x^4)} + \sqrt[3]{(x^6 - 2ax^5 + a^2 x^4)} \right )} =$ (dividing numerator and denominator by $x^2 = \sqrt[3]{x^6}$ ) $= \lim_{x \to \infty} \frac{2a}{\left ( \sqrt[3]{(1 + \frac{2a}{x} + \frac{a^2}{x^2})} + \sqrt[3]{(1 - \frac{a^2}{x^2})} + \sqrt[3]{(1 - \frac{2a}{x} + \frac{a^2}{ x^2})} \right )} = \frac{2a}{3}.$ Substituing $a = 2018$ we get the result.

$\boxed{2}$ If $|x| < 1$ , $\displaystyle \quad \sqrt[3]{1 + x} = \sum_{n = 0}^\infty {1/3 \choose n} x^n$ . $\quad \left ( {1/3 \choose n} = \frac{(1/3) \cdot (1/3 - 1) \cdot \ldots \cdot (1/3 - (n - 1))}{n!} \right ) \quad$ . Due to $\frac{2018}{x} < 1$ as $x \to \infty$ , $\displaystyle \lim_{x \to \infty} \left ( \sqrt[3]{x^3 + 2018x^2} - \sqrt[3]{x^3 - 2018x^2} \right ) = \lim_{x \to \infty} x\left ( \sqrt[3]{1 + \frac{2018}{x}} - \sqrt[3]{1 - \frac{2018}{x}} \right ) = \lim_{x \to \infty} x \left ( (1 + \frac{1}{3} \cdot \frac{2018}{x} + o(\frac{2018}{x})) - (1 - \frac{1}{3} \cdot \frac{2018}{x} + o(\frac{2018}{x})) \right )$ $= \frac{2}{3} \cdot 2018$

$\begin{aligned}& \text{L} = \displaystyle\lim_{x\to\infty} \left[({x^3+2018x^2})^{\frac{1}{3}} - ({x^3-2018x^2})^{\frac{1}{3}}\right]\\& =\displaystyle\lim_{x\to\infty}\left[x({1+\frac{2018}{x}})^{\frac{1}{3}} - x({1-\frac{2018}{x}})^{\frac{1}{3}}\right] \\& = \displaystyle\lim_{x\to\infty}\left[\frac{(1+\frac{2018}{x})^{\frac{1}{3}}-(1-\frac{2018}{x})^{\frac{1}{3}}}{\frac{1}{x}}\right]\end{aligned}$ On placing the value of $x=0$ then $\text{L} =\frac{0}{0}$ form so now applying L-Hopital's rules We get

$\begin{aligned}& \text{L} = \displaystyle\lim_{x\to\infty}\left[\frac{\frac{-2018}{3x^2}(1+\frac{2018}{x})^{-\frac{2}{3}}-\frac{2018}{3x^2}(1-\frac{2018}{x})^{-\frac{2}{3}}}{-\frac{1}{x^2}}\right]\\& = \frac{2018}{3} \displaystyle\lim_{x\to\infty}\left[-\frac{1}{x^2}\frac{(1+\frac{2018}{x})^{-\frac{2}{3}}+ (1-\frac{2018}{x})^{-\frac{2}{3}}}{-\frac{1}{x^2}}\right]\\&= \frac{2018}{3}\displaystyle\lim_{x\to\infty}\left[(1+\frac{2018}{x})^{-\frac{2}{3}}+ (1-\frac{2018}{x})^{-\frac{2}{3}}\right] \end{aligned}$ $\begin{aligned}\text{L} = \frac{2}{3}. 2018 = \frac{A}{B}.2018\end{aligned}$ Hence the value of $10A - B =\boxed{17}$