Let f(x) = frac{x}{(1+x^n)^{1/n}},x in R - {- | vedclass

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(C) Given $f(x) = \frac{x}{(1+x^n)^{1/n}}$.We observe the pattern for composition:$f(f(x)) = \frac{f(x)}{(1+(f(x))^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{

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(C) Given $f(x) = \frac{x}{(1+x^n)^{1/n}}$.We observe the pattern for composition:$f(f(x)) = \frac{f(x)}{(1+(f(x))^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{(1 + x^n/(1+x^n))^{1/n}} = \frac{x}{(1+x^n+x^n)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$.By induction,$f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$.We need to evaluate $I = \lim_{n 	o \infty} \int_0^1 x^{n-2} \frac{x}{(1+nx^n)^{1/n}} dx = \lim_{n 	o \infty} \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}} dx$.Let $t = 1 + nx^n$,then $dt = n^2 x^{n-1} dx$,so $x^{n-1} dx = \frac{dt}{n^2}$.As $x 	o 0, t 	o 1$ and as $x 	o 1, t 	o 1+n$.$I = \lim_{n 	o \infty} \frac{1}{n^2} \int_1^{1+n} t^{-1/n} dt = \lim_{n 	o \infty} \frac{1}{n^2} \left[ \frac{t^{1-1/n}}{1-1/n} ight]_1^{1+n}$.$I = \lim_{n 	o \infty} \frac{1}{n^2} \frac{n}{n-1} ((1+n)^{(n-1)/n} - 1) = \lim_{n 	o \infty} \frac{1}{n(n-1)} ((1+n)^{1-1/n} - 1)$.Since $(1+n)^{1-1/n} \approx n$,the expression behaves like $\frac{n}{n^2} 	o 0$ as $n 	o \infty$.

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

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