I couldn't find the right path for this limit to solve. $$\lim_{n\to\infty}\left(\dfrac{f(1)+f(2)+...+f(n)}{n}\right)^n$$
$$f:\mathbb{R}\to\mathbb{R}, f(x)=\sqrt[3]{x^3+3x^2+2x+1}-\sqrt[3]{x^3-x+1}$$
I know that we have the indeterminate $1^{\infty}$, but apart from that...I didn't get very far
As N.S.John says in the comments, we can write $$\sqrt [3]{x^3+3x^2+2x+1} =\sqrt [3]{(x+1)^3-(x+1)+1}$$ and $$\sqrt [3]{x^3-x+1} =\sqrt [3]{(x-1+1)^3-(x-1+1)+1} $$ Thus $$f (1)=\sqrt [3]{2^3-2+1}-\sqrt [3]{1^3-1+1} $$ $$f (2)=\sqrt [3]{3^3-3+1}-\sqrt[3]{2^3-2+1}$$ $$\vdots $$ $$f (n)=\sqrt [3]{(n+1)^3-(n+1)+1}-\sqrt [3]{n^3-n+1} $$ $$\Rightarrow \sum_{i=1}^{n} f (i) = \sqrt [3]{(n+1)^3-(n+1)+1}-\sqrt [3]{1} $$ Can you take it from here? The answer is $\boxed {1}$ assuming $\lim_{n \to \infty}$. Hope it helps.