Find: $\displaystyle\lim_{x\to\infty}x^{\frac{5}{3}}\left[\big(x+\sin{({1}/{x})}\big)^{\frac{1}{3}}-x^{\frac{1}{3}}\right]$ $"\infty * 0"$.
The limit might not exist, but than I have to prove it.
 A: Some extended hint. Pull out a factor $x$ from the bracket $(x+\sin(1/x))$ to obtain
$$
x^{5/3} [x^{1/3} ((1+\sin(1/x)/x)^{1/3}-1)]\ .
$$
Then expand $$
\frac{\sin(1/x)}{x}\sim \frac{1}{x^2}
$$
for $x\to\infty$. [You can show it by changing variable $x\to 1/t$ and consider the expression $t\sin(t)=t^2 \left(\frac{\sin t}{t}\right)$ for $t\to 0$]. Therefore, you should try to compute
$$
\lim_{x\to\infty} x^2 \left[\left(1+\frac{1}{x^2}\right)^{1/3}-1\right]\ .
$$
A: Set $1/x=h$ to find
$$F=\lim_{h\to0}\dfrac{(1+h\cdot\sin h)^{1/3}-1}{h^2}$$
Using $a^3-b^3=(a-b)(a^2+ab+b^2),$
$$F=\lim_{h\to0}\dfrac{(1+h\cdot\sin h)-1}{h^2}\cdot\dfrac1{\lim_{h\to0}\{\sum_{r=0}^2(1+h\cdot\sin h)^{r/3}\}}$$
Can you take it from here?
A: $$x+\sin{(\frac{1}{x})} = _{\infty}x+\frac{1}{x}+o(\frac{1}{x}) $$
So :
$$(x+\frac{1}{x})^{\frac{1}{3}}=x^{\frac{1}{3}}(1+{\frac{1}{x^2}})=_{\infty}x^{\frac{1}{3}}(1+{\frac{1}{3x^2}})+o(\frac{1}{x^{5/3}})=_{\infty}x^{\frac{1}{3}}+\frac{1}{3*x^{5/3}}+o(\frac{1}{x^{5/3}}) $$
Finally :
$$x^{\frac{5}{3}}[(x+\sin{(\frac{1}{x})})^{\frac{1}{3}}-x^{\frac{1}{3}}]=_{\infty}x^{\frac{5}{3}}[x^{\frac{1}{3}}+\frac{1}{3*x^{5/3}}-x^{\frac{1}{3}}+o(\frac{1}{x^{5/3}})]=_{\infty}\frac{1}{3}+o(1)$$
We can conclude : 
$$\lim_{x\to\infty}x^{\frac{5}{3}}[(x+\sin{(\frac{1}{x})})^{\frac{1}{3}}-x^{\frac{1}{3}}]=\frac{1}{3}$$
