Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$ Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$
I converted it to $\lim \limits_{t \to 0}$ and tried using L'Hospital's rule and I got it after differentiating twice. It looks pretty bad. Please give an elegant solution. Thank you.

Observe that $(n+1)^3=n^3+3n^2+3n+1$, $(n-1)^2=n^2-2n+1$, $2n=(n+1)+(n-1)$
$$\lim \limits_{n \to \infty}n((n+1)^3-n)^{1/3}-(n+1))\\= \lim \limits_{n \to \infty}\frac{-n^2}{((n+1)^3-n)^{2/3}+(n+1)((n+1)^3-n)^{1/3}+(n+1)^2}\\=\frac{-1}{3}$$
$$\lim \limits_{n \to \infty}n((n-1)^2+2)^{1/2}-(n-1))=\lim \limits_{n \to \infty}\frac{2n}{((n-1)^2+2)^{1/2}+(n-1)}=1$$
so $k=1-\frac{1}{3}=\frac{2}{3}$
