$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $ fine limit :
$$ \lim_{x \to  \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$
such that : $$ a \in (0,2)$$
and :
$[x]: \ \ $  floor function 
My Try :
$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}$$
$$\ln f(x)=(x-[x])\ln(\sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )$$
Now ?please help
 A: The expression under limit operation say $F(x) $ is of the form $\{f(x) \} ^{g(x)} $ and clearly $f(x) >0$ for all $x>0$ and $0\leq g(x) <1$ for all $x>0$. It is now clear that $F(x) $ lies between $1$ and $f(x) $ and it takes the value $1$ when $x$ is a positive integer. Moreover if $x$ is near a positive integer and less than it then $F(x) $ is near $f(x) $. Hence it is clear that if $F(x) $ tends to a limit as $x\to\infty$ then it must be $1$ and moreover this will happen if and only if $f(x) \to 1$ as $x\to\infty$.
Now it is easy to analyze $f(x) $ which is of the form $p-q$. Multiplying it by $(p^{2}+pq+q^{2})$ and dividing it by the same quantity we see that $f(x) $ is expressed as a fraction with numerator $p^{3}-q^{3}=3x^{a}$. Further if divide the numerator and denominator of $f(x) $ by $x^{4/3}$ then the denominator tends to $3$ and the numerator is $3x^{a-4/3}$. It follows that $f(x) \to 1$ if and only if $a=4/3$. Hence the desired limit is equal to $1$ if $a=4/3$ and the limit does not exist if $a\neq 1$.
A: $\begin{array}\\
( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}
&=(x^{2/3} \sqrt[3]{1+4 x^{a-2}} - x^{2/3}\sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (\sqrt[3]{1+4 x^{a-2}} - \sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (1+4 x^{a-2}/3+O(x^{a-3}) - (1+ x^{a-2}/3+O(x^{a-3}))  )^{x-[x]}\\
&=x^{2(x-[x])/3} ( x^{a-2}+O(x^{a-3} ))^{x-[x]}\\
&=x^{(x-[x])(2/3+a-2)} ( 1+O(x^{-1} ))^{x-[x]}\\
&=x^{(x-[x])(a-4/3)} ( 1+O(x^{-1} ))^{x-[x]}\\
\end{array}
$
As Paramanand Singh's answer states,
if $a = 4/3$,
this is
$( 1+O(x^{-1} ))^{x-[x]}
$
which goes to $1$.
If $a \ne 4/3$,
this goes to
$x^{(x-[x])(a-4/3)}
$.
This goes from
$1$
(when $x = [x]$)
to
$x^{a-4/3}$
(when
$x-[x] \approx 1$).
Therefore,
when $a \ne 4/3$,
the limit does not exist.
