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?
 A: First I would factor $x^{1/3}$ out of the second term:
$$
x^{5/3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)
 = x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right)
$$
Then I would substitute $t = \frac{1}{x}$:
$$
    \lim_{x\to\infty}x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right)
   = \lim_{t\to 0^+} \frac{(1+t \sin t)^{1/3} - 1}{t^2}
$$
At this point we could use Taylor's Theorem, or the Binomial series, or L'Hôpital's rule.  The limit of the quotient of the derivatives is:
\begin{align*}
    \lim_{t\to 0^+} \frac{\frac{1}{3}(1+t\sin t)^{-2/3}(t \cos t + \sin t)}{2t}
    &= \frac{1}{6}\cdot 1 \cdot \lim_{t\to 0^+} \left(\cos t + \frac{\sin t}{t}\right)
    \\&= \frac{1}{6}(1+1) = \frac{1}{3}
\end{align*}
A: $$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)=  \lim_{x\to\infty}x^2\left(\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1\right)=\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1}{\frac{1}{x^2}}$$
Now using L'Hopital we get
$$\lim_{x\to\infty}\frac{\frac{1}{3}\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{-2}{3}\left(\frac{-1}{x^2}\right)\left( \sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right) \right)}{\frac{-2}{x^3}}=\frac{1}{6}\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right)}{\frac{1}{x}}=\\ \frac{1}{6}\lim_{x\to\infty}\frac{\frac{-2}{x^2}\cos\left(\frac{1}{x}\right)-\frac{-1}{x^3}\sin\left(\frac{1}{x}\right)}{\frac{-1}{x^2}}=\frac{1}{6}\times2=\frac{1}{3}$$
A: I think using the difference of cubes should work:$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)=\\
\lim_{x\to\infty}\frac{x^\frac{5}{3}\sin\frac{1}{x}}{\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{2}{3}+\left(x+\sin\frac{1}{x}\right)^{\frac{1}{3}}x^\frac{1}{3}+x^\frac{2}{3}}=\\
\lim_{x\to\infty}\frac{x^\frac{5}{3}x^{-1}}{\left(x+x^{-1}\right)^\frac{2}{3}+\left(x+x^{-1}\right)^\frac{1}{3}x^\frac{1}{3}+x^\frac{2}{3}}=\\
\lim_{x\to\infty}\frac{x^\frac{2}{3}}{x^\frac{2}{3}+x^\frac{2}{3}+x^\frac{2}{3}}=\frac{1}{3}
$$
A: 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 $x \to +\infty$
