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?

Always try to simplify algebraic calculations in a manner which reduces typing effort and visual clutter.
Clearly we can take $n$ common from both terms in denominator and hence denominator can be written as $n(a-b) $ where both $a, b$ tend to $1$. Further we can see that $a^3,b^2$ are radical free and hence we have $$n(a-b) =n(a-1-(b-1))=n\left((a^3-1)\cdot\frac{a-1}{a^3-1}-(b^2-1)\cdot\frac{b-1}{b^2-1}\right)\tag{1}$$ Just note that $$n(a^3-1)=n\left(\frac{1}{n^2}+\frac {1}{n^3}\right)\to 0$$ and $$n(b^2-1)=n\left(-\frac{1}{n}+\frac{2}{n^2}\right)\to - 1$$ It now follows from equation $(1)$ that denominator $n(a-b) $ tends to $$0\cdot\frac{1}{3}-(-1)\cdot\frac{1}{2}=\frac{1}{2}$$ and the expression under limit thus tends to $2$.
