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?
 A: 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
A: fast solution,as another idea$$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+0n^2+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{(n+0)^3-0+0n^2+n+1}-\sqrt{(n-\frac12)^2-\frac14+2}}=\\
\lim_{n \to \infty}\frac{1}{n-(n-\frac12)}=\\2$$
REMARK:
$$n^3+an^2+bn+c=\\(n+\frac a3)^3-(3n^2.\frac a3+3n(\frac  a3)^2+(\frac a3)^3)+bn+c\\(n\to \infty) \implies n^3+an^2+bn+c\sim (n+\frac a3)^3 $$ so $$n^3+n+1=(n+0)^3-0^3+n+1$$ also for $n^2+an+b=(n+\frac a2)^2-(\frac a2)^2+c$
A: 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 $2$.
A: For large $n$, $(1+n^{-2}+n^{-3})^{1/3}\in 1+\tfrac13n^{-2}+O(n^{-2})\subseteq 1+o(n^{-1})$, so$$\begin{align}\frac{n^{-1}}{(1+n^{-2}+n^{-3})^{1/3}-(1-n^{-1}+2n^{-2})^{1/2}}&\in\frac{n^{-1}}{1+o(n^{-1})-1+\tfrac12n^{-1}+o(n^{-1})}\\&=\frac{n^{-1}}{\tfrac12n^{-1}+o(n^{-1})}\\&\stackrel{n\to\infty}{\sim}2.\end{align}$$
