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.
Setting $1/n=h,$ $F=\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$
$=\lim_{h\to0^+}\dfrac{(1+3h+2h^2+h^3)^{1/3}+(1-2h+3h^2)^{1/2}-2}{h^2}$
Using Binomial series,
$(1+3h+2h^2+h^3)^{1/3}=\{1+h(3+2h+h^2)\}^{1/3}$ $=1+1/3h(3+2h+h^2)+1/3(1/3-1)\dfrac{\{h(3+2h+h^2)\}^2}{2!}+O(h^3)$ $=1+h+h^2\left(\dfrac23-\dfrac29\cdot\dfrac92\right)+O(h^3)=1+h-h^2/3+O(h^3)$
$(1-2h+3h^2)^{1/2}=\{1+h(3h-2)\}^{1/2}$ $=1+1/2h(3h-2)+1/2(1/2-1)\dfrac{\{h(3h-2)\}^2}{2!}+O(h^3)$ $=1-h+h^2\left(\dfrac32-\dfrac14\cdot\dfrac92\right)+O(h^3)$ $=1-h-3h^2/8+O(h^3)$
Can you take it from here?
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 $k=1-\frac{1}{3}=\frac{2}{3}$
Start with $$\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n] = \lim \limits_{n \to \infty} n^2\left(\left(1+\frac 3n+\frac 2{n^2}+\frac 1{n^3}\right)^{1/3}+\left(1- \frac 2n+ \frac 3{n^2}\right)^{1/2}-2\right), $$
then use the expansion $(1+x)^a = 1 + a x + \frac{a(a-1)}2 x^2 + o(x^2)$ to expand the two remaining terms
