$\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$ $\lim_{n\to+\infty}  {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$
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 ?

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 $\sqrt[3]{1+mh}-1=r\implies mh=3r+3r^3+r^3$ $$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+mh}-1}h=\lim_{r\to0^+}\dfrac{rm}{3r+3r^3+r^3}=\frac m3$$
