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 ?

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