Let $$f(x) = \dfrac{x}{(1+x^n)^{1/n}}$$, $$x \in \mathbb{R} - \{-1\}$$, $$n \in \mathbb{N}$$, $$n > 2$$. If $$f^n(x) = (f \circ f \circ f \ldots$$ upto n times$$(x)$$, then $$\lim_{n \to \infty} \int_0^1 x^{n-2}(f^n(x))dx$$ is equal to ______.

Given $$f(x) = \dfrac{x}{(1+x^n)^{1/n}}$$, $$n \in \mathbb{N}$$, $$n > 2$$.

Finding the $$n$$-fold composition $$f^n(x)$$:

$$f(x) = \frac{x}{(1+x^n)^{1/n}}$$, so $$[f(x)]^n = \frac{x^n}{1+x^n}$$

$$f(f(x)) = \frac{f(x)}{(1+[f(x)]^n)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$$

By induction: $$f^k(x) = \frac{x}{(1+kx^n)^{1/n}}$$

Therefore $$f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$$.

Computing the integral:

$$I_n = \int_0^1 x^{n-2} \cdot \frac{x}{(1+nx^n)^{1/n}}\,dx = \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}}\,dx$$

Substituting $$u = 1 + nx^n$$, $$du = n^2 x^{n-1}\,dx$$:

$$I_n = \frac{1}{n^2}\int_1^{1+n} u^{-1/n}\,du = \frac{1}{n^2} \cdot \frac{n}{n-1}\left[(1+n)^{(n-1)/n} - 1\right]$$

$$= \frac{1}{n(n-1)}\left[(n+1)^{(n-1)/n} - 1\right]$$

Taking the limit as $$n \to \infty$$:

$$(n+1)^{(n-1)/n} = (n+1) \cdot (n+1)^{-1/n} \to (n+1) \cdot 1 = n+1$$

$$\lim_{n \to \infty} I_n = \lim_{n \to \infty} \frac{n+1-1}{n(n-1)} = \lim_{n \to \infty} \frac{n}{n(n-1)} = \lim_{n \to \infty} \frac{1}{n-1} = 0$$

The answer is $$0$$.