<s>[INST] $ \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
 [/INST] $\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.
</s>