How to find this limit : $x\sin{f(x)}$

Question

How to find the limit $$\lim_{x\to\infty}x\sin f(x)$$ where $$f(x)=\left(\sqrt[3]{x^3+4x^2}-\sqrt[3]{x^3+x^2}\right)\pi\ ?$$ Is it possible to solve without L'Hospital's rule ?

Answer 1

Setting $x=u^{-1}$, we have \begin{eqnarray} x\sin f(x)&=&\frac{1}{u}\sin\left[\pi\left(\sqrt[3]{u^{-3}+4u^{-2}}-\sqrt[3]{u^{-3}+u^{-2}}\right)\right]=\frac1u\sin\left[\pi\frac{\sqrt[3]{1+4u}-\sqrt[3]{1+u}}{u}\right]\\ &=&\frac1u\sin\left[\pi\frac{(1+\frac{4}{3}u-\frac{16}{9}u^2)-(1+\frac{u}{3}-\frac{u^2}{9})+o(u^2)}{u}\right]\\ &=&\frac1u\sin\left[\pi\frac{u-\frac{5}{3}u^2+o(u^2)}{u}\right]\\ &=&\frac1u\sin\left[\pi-\frac{5\pi}{3}u+o(u)\right]\\ &=&\frac1u\sin\left[\frac{5\pi}{3}u+o(u)\right]\\ &=&\frac1u\left[\frac{5\pi}{3}u+o(u)\right]\\ &=&\frac{5\pi}{3}+o(1). \end{eqnarray} It follows that $$ \lim_{x\to\infty}x\sin f(x)=\lim_{u\to0}\frac1u\sin f\left(\frac1u\right)=\lim_{u\to0}\left[\frac{5\pi}{3}+o(1)\right]=\frac{5\pi}{3}. $$

Answer 2

Write

$\begin{array}\\ (x^3+ax^2)^{1/3} &=x(1+a/x)^{1/3}\\ &=x(1+(a/3x)+(1/3)(-2/3)(a/x)^2/2 + O(1/x^3))\\ &=x+(a/3)-(a^2/9x) + O(1/x^2))\\ \end{array} $

so

$\begin{array}\\ (x^3+ax^2)^{1/3}-(x^3+bx^2)^{1/3} &=(x+(a/3)-(a^2/9x) + O(1/x^2)))-(x+(b/3)-(b^2/9x) + O(1/x^2)))\\ &=(a-b)/3-(a^2-b^2)/(9x) + O(1/x^2)\\ \end{array} $

Therefore, setting $a=4, b=1$,

$\begin{array}\\ \sin f(x) &= \sin(\pi(1-15/(9x)+O(1/x^2)))\\ &=\sin(\pi(5/(3x)+O(1/x^2)))\\ &=\pi(5/(3x)+O(1/x^2)\\ \end{array} $

so

$x \sin f(x) =5\pi/3+O(1/x) .$