I have limit:
$\lim_{n\to\infty} (\sqrt[3]{n^{48}+n} - \sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$\dfrac{(\sqrt[3]{n^{48}+n})^2 + (\sqrt[3]{n^{48}+n})(\sqrt[3]{n^{48}+n^2})+ (\sqrt[3]{n^{48}+n^2})^2)}{(\sqrt[3]{n^{48}+n})^2 + (\sqrt[3]{n^{48}+n})(\sqrt[3]{n^{48}+n^2})+ (\sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$\dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(\sqrt[3]{1n^{48}+n})^2 + (\sqrt[3]{1n^{48}+n})(\sqrt[3]{1n^{48}+n^2})+ (\sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$\dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(\sqrt[3]{1n^{48}+n})^2 + (\sqrt[3]{1n^{48}+n})(\sqrt[3]{1n^{48}+n^2})+ (\sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\\ =((n^{36}+36n^{33}+594n^{30}+\cdots)-(n^{36}+36n^{33}+576n^{30}+\cdots))\\ =18n^{30}+\cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(\sqrt[3]{n^{48}+n}-\sqrt[3]{n^{48}+n^{2}})\\ =n^{16}(\sqrt[3]{1+n^{-47}}-\sqrt[3]{1+n^{-46}})\\ =n^{16}((1+\frac{1}{3}n^{-47}+\cdots)-(1+\frac{1}{3}n^{-46}+\cdots))\\ =n^{16}(-\frac{1}{3}n^{-46}+\frac{1}{3}n^{-47}+\cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+\cdots)$$
which equals $-6$ in the limit.
Hope that helps