Let f(x) = x (1 + x ^ n ) ^ (1)/(n) , x ∈ ℝ - \ - 1\ , n ∈ ℕ, n > 2...

Given,

$f\left(x\right) = \dfrac{x}{{\left(1 + {x}^{n}\right)}^{\dfrac{1}{n}}}, x \in \mathbb{R} - \left\{- 1\right\}, n \in \mathbb{N}, n > 2$,

${f}^{n}\left(x\right) = ( fofof . . . .$ upto $n$ times) \left(x\right)$,</p><p>Now finding$f\left(f\left(x\right)\right) = \dfrac{f\left(x\right)}{{\left[1 + {\left(f\left(x\right)\right)}^{n}\right]}^{\dfrac{1}{n}}}$</p><p>$⇒ f\left(f\left(x\right)\right) = \dfrac{\dfrac{x}{{\left(1 + {x}^{n}\right)}^{\dfrac{1}{n}}}}{{\left[1 + {\left(\dfrac{x}{{\left(1 + {x}^{n}\right)}^{\dfrac{1}{n}}}\right)}^{n}\right]}^{\dfrac{1}{n}}}$</p><p>$⇒ f\left(f\left(x\right)\right) = \dfrac{x}{{\left[1 + 2{x}^{n}\right]}^{\dfrac{1}{n}}}$</p><p>Similarly,${f}^{n}\left(x\right) = \dfrac{x}{{\left(1 + n{x}^{n}\right)}^{\dfrac{1}{n}}}$</p><p>Now solving the integral we get,</p><p>$I = {∫}_{0}^{1}\dfrac{{x}^{n - 1}}{{\left(1 + n{x}^{n}\right)}^{\dfrac{1}{n}}}dx$</p><p>Now let,$1 + n{x}^{n} = {t}^{n}$</p><p>$⇒ {n}^{2}{x}^{n - 1}dx = n{t}^{n - 1}dt$</p><p>$I = {∫}_{1}^{\left(1 + n\right)\dfrac{1}{n}}\dfrac{\dfrac{1}{n}{t}^{n - 1}dt}{t}$</p><p>$⇒ I = {\left[\dfrac{1}{n}\dfrac{{t}^{n - 1}}{n - 1}\right]}_{1}^{\left(1 + n\right)\dfrac{1}{n}}$</p><p>$⇒ I = \dfrac{1}{n\left(n - 1\right)}\left({\left(1 + n\right)}^{1 - \dfrac{1}{n}} - 1\right)$</p><p>Now solving the limit we get,</p><p>$\lim_{n \to \infty}\dfrac{{\left(1 + n\right)}^{1 - \dfrac{1}{n}} - 1}{n\left(n - 1\right)}$</p><p>Now let$n = \dfrac{1}{h}$so the limit changes to,</p><p>$\lim_{h \to 0}\dfrac{{\left(1 + \dfrac{1}{h}\right)}^{1 - h} - 1}{\dfrac{1}{h}\left(\dfrac{1}{h} - 1\right)}$</p><p>$= \lim_{h \to 0}\dfrac{\left(1 + \dfrac{1}{h}\left(1 - h\right)\right) - 1}{\dfrac{1}{h}\left(\dfrac{1}{h} - 1\right)}$</p><p>$= \lim_{h \to 0}\dfrac{\left(1 - h\right)}{\left(\dfrac{1}{h} - 1\right)}$</p><p>$= \lim_{h \to 0}\dfrac{h\left(1 - h\right)}{\left(1 - h\right)}$</p><p>$= \lim_{h \to 0}h = 0