$\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$ $\lim_{n\to+\infty}  {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$
I used an extension of the appropriate 1 (A ^ 2 - AB + B ^ 2) / (A ^ 2 - AB + B ^ 2). I got here
$\lim_{n\to+\infty} {n^{(4/3)}} \Big(4/ (({n^2 +1} )^{(2/3)} + (n^4-2n^2-3)^{(1/3)}+(3-n^2)^{(2/3)})\Big)$
I'll blame the leading term in the denominator -> n ^ (4/3)
$\lim_{n\to+\infty} \Big(4/ (({1+0} )^{(2/3)} + (1+0+0)^{(1/3)}+(0-1)^{(2/3)})\Big)$
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$(0-1)^{(2/3)}$
I care about this term in the denominator because it is equal to -1, but due to the exponentiation it is always positive -> How will it affect the final result ?
Final result is 4/3 op 4/1 ?
 A: The signs are a bit less confusing if we write the sum of two cube roots, one of them negative, as a difference of two cube roots, both of them positive. In terms of $a:=\sqrt[3]{n^2+1},\,b:=\sqrt[3]{n^2-3}$ (so in your notation $A=a,\,B=-b$), $a^3-b^3=4$ so you want$$\lim_{n\to\infty}\frac{4n^{4/3}}{a^2+ab+b^2}.$$The denominator's three terms are each asymptotic to $n^{4/3}$, which should answer your question. If you don't like "is asymptotic to" arguments, @Koro's tip is to define$$\alpha:=\frac{a}{n^{2/3}}=\sqrt[3]{1+1/n^2},\,\beta:=\frac{a}{n^{2/3}}=\sqrt[3]{1-3/n^2},$$in terms of which we want$$\lim_{n\to\infty}\frac{4}{\alpha^2+\alpha\beta+\beta^2}.$$The limits of $\alpha,\,\beta$ make this simple. Of course, if you don't take my sign-changing advice, you'll do the equivalent of calculating $1^2-(1)(-1)+(-1)^2$ instead of $1^2\times3$.
A: Pedestrian approach:
$1):n^{2/3}(1+1/n^2)^{1/3}=n^{2/3}(1+(1/3)(1/n^2)-.....);$
$2):-n^{2/3}(1-3(1/n^2))^{1/3}=-n^{2/3}(1-(1/n^2) +........);$
Add $1)+2):$ $n^{2/3}((4/3)(1/n^2)+......);$
Can you finish?
Used: Binomial expansion
A: WLOG $\dfrac1{n^2}=h$  to find  $\sqrt[3]{3-n^2}=\sqrt[3]{\dfrac{3h-1}h}=\dfrac{\sqrt[3]{3h-1}}{h^{1/3}}=-\dfrac{\sqrt[3]{1-3h}}{h^{1/3}}$
So, the limit in question reduces to
$$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-\sqrt[3]{1-3h}}h=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt[3]{1-3h}-1}h$$
Now for set $\sqrt[3]{1+mh}-1=r\implies mh=3r+3r^3+r^3$ $$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+mh}-1}h=\lim_{r\to0^+}\dfrac{rm}{3r+3r^3+r^3}=\frac m3$$
