An indeterminate limit form of infinity/infinity I am trying to solve the limit:
$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$
I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$
But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?

You have that
$$
\begin{gathered}
  \mathop {\lim }\limits_{x \to  + \infty } x^{\frac{5}
{3}} \left[ {\left( {x + \sin \left( {\frac{1}
{x}} \right)} \right)^{\frac{1}
{3}}  - x^{\frac{1}
{3}} } \right] =  \hfill \\
   \hfill \\
  \mathop {\lim }\limits_{x \to  + \infty } x^{\frac{5}
{3}} \left[ {x^{\frac{1}
{3}} \left( {1 + \frac{1}
{x}\sin \left( {\frac{1}
{x}} \right)} \right)^{\frac{1}
{3}}  - 1} \right] =  \hfill \\
   \hfill \\
   = \mathop {\lim }\limits_{x \to  + \infty } x^2 \left[ {\frac{1}
{{3x}}\sin \left( {\frac{1}
{x}} \right)} \right] =  \hfill \\
   \hfill \\
   = \frac{1}
{3}\mathop {\lim }\limits_{x \to  + \infty } x\left[ {\sin \left( {\frac{1}
{x}} \right)} \right] = \frac{1}
{3}\mathop {\lim }\limits_{x \to  + \infty } \left[ {\frac{{\sin \left( {\frac{1}
{x}} \right)}}
{{\frac{1}
{x}}}} \right] = \frac{1}
{3} \hfill \\ 
\end{gathered} 
$$
where we used  the fact that
$$
\left( {1 + f\left( x \right)} \right)^\alpha   - 1 \sim \alpha f(x)\,\,\,\,\,\,\left( {x \to x_0 } \right)
$$
provided 
$$
\mathop {\lim }\limits_{x \to x_0 } f(x) = 0
$$
and the fact that
$$
\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}
{t} = 1
$$
with $t=1/x$ as $x \to +\infty$
