How to find this limit : $x\sin{f(x)}$ 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 ?
 A: 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
$x \sin f(x)
=5\pi/3+O(1/x)
.$
A: 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
$$
\lim_{x\to\infty}x\sin f(x)=\lim_{u\to0}\frac1u\sin f\left(\frac1u\right)=\lim_{u\to0}\left[\frac{5\pi}{3}+o(1)\right]=\frac{5\pi}{3}.
$$
