<s>[INST] How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital? $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$
$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:
$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$ I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?
 [/INST] binomial theorem for rational exponents:
(1+n)^(1/3) = 1 + n/3 + ...
(1+n)^(1/2) = 1 + n/2 + ...
s1= (n^3 + n - 1)^(1/3) = [n^3 (1+ 1/n^2 + ...)]^(1/3) = n (1+ 1/(3 n^2) ... )
s2= (n^2 -n  + 2)^(1/2) = [n^2 (1- 1/n^2 + ...)]^(1/2) = n (1 -1/(2n) ... )
s1-s2 = 1/(3n) + 1/2 ...
lim  1/(s1-s2) = 2
n->infinity
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