Let f x = x 1 + x n 1 n ,   x ∈ ℝ - - 1 ,   n ∈ ℕ ,   n > 2 . If…

Given,

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

$f^n\left(x\right)=(fofof....$ upto $n$ times) $\left(x\right)$,

Now finding $f\left(f\left(x\right)\right)=\frac{f\left(x\right)}{{\left[1+{\left(f\left(x\right)\right)}^n\right]}^{{\displaystyle \frac{1}{n}}}}$

$\Rightarrow f\left(f\left(x\right)\right)=\frac{\frac{x}{{\left(1+x^n\right)}^{{\displaystyle \frac{1}{n}}}}}{{\left[1+{\left(\frac{x}{{\left(1+x^n\right)}^{{\displaystyle \frac{1}{n}}}}\right)}^n\right]}^{{\displaystyle \frac{1}{n}}}}$

$\Rightarrow f\left(f\left(x\right)\right)=\frac{x}{{\left[1+2x^n\right]}^{{\displaystyle \frac{1}{n}}}}$

Similarly, $f^n\left(x\right)=\frac{x}{{\left(1+n x^n\right)}^{{\displaystyle \frac{1}{n}}}}$

Now solving the integral we get,

$I={\int }_0^1\frac{x^{n-1}}{{\left(1+n{x}^n\right)}^{{\displaystyle \frac{1}{n}}}}dx$

Now let, $1+n{x}^n=t^n$

$\Rightarrow n^2{x}^{n-1}dx=n{t}^{n-1} dt$

$I={\int }_1^{{\left(1+n\right)}^{\frac{1}{n}}} \frac{{\displaystyle \frac{1}{n}}{t}^{n-1} dt}{t}$

$\Rightarrow I={\left[\frac{1}{n}\frac{t^{n-1}}{n-1}\right]}_1^{{\left(1+n\right)}^{\frac{1}{n}}}$

$\Rightarrow I=\frac{1}{n\left(n-1\right)}\left({\left(1+n\right)}^{1-\frac{1}{n}}-1\right)$

Now solving the limit we get,

$\underset{n\to \infty }{\lim }\frac{{\left(1+n\right)}^{1-{\displaystyle \frac{1}{n}}}-1}{n\left(n-1\right)}$

Now let $n=\frac{1}{h}$ so the limit changes to,

$\underset{h\to 0}{\lim }\frac{{\left(1+\frac{1}{h}\right)}^{1-{\displaystyle h}}-1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}$

$=\underset{h\to 0}{\lim }\frac{\left(1+\frac{1}{h}\left(1-h\right)\right)-1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}$

$=\underset{h\to 0}{\lim }\frac{\left(1-h\right)}{\left(\frac{1}{h}-1\right)}$

$=\underset{h\to 0}{\lim }\frac{h\left(1-h\right)}{\left(1-h\right)}$

$=\underset{h\to 0}{\lim }h=0$