$ \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

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 $a\neq 1$.
