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

If $f^{n}(x)=\left(f \circ f \circ f \ldots .\right.$. upto $n$ times) $(x)$, then

$$\lim _\limits{n \rightarrow \infty} \int_\limits{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x$$ is equal to ____________.

$$ \begin{aligned} & \text { We have, } f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} \\\\ & \therefore f(f(x))=\frac{f(x)}{\left(1+\left[f(x)^n\right]^{1 / n}\right.}\\\\ &=\frac{\frac{x}{\left(1+x^n\right)^{1 / n}}}{\left(1+\frac{x^n}{1+x^n}\right)^{1 / n}}\\\\ &=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\\\ & f(f(f(x)))=\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\\\ & \text { Similarly, } f^n(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \end{aligned} $$ <br/><br/>Now, $\lim\limits_{n \rightarrow \infty} \int_0^1 x^{n-2}\left(f^n(x)\right) d x$ <br/><br/>$$ \begin{aligned} & =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 x^{n-2} \frac{x}{\left(1+n x^n\right)^{1 / n}} d x \\\\ & =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 \frac{x^{n-1}}{\left(1+n x^n\right)^{1 / n}} d x \end{aligned} $$ <br/><br/>Let $1+n x^n=t$ <br/><br/>$\Rightarrow n^2 x^{n-1} d x=d t$ <br/><br/>When, $x=0$, then $t=1$ <br/><br/>When, $x=1$, then $t=1+n$ <br/><br/>$\begin{aligned} & =\lim _{n \rightarrow \infty} \int_1^{1+n} \frac{d t}{n^2(t)^{1 / n}} =\lim _{n \rightarrow \infty} \frac{1}{n^2} \int_1^{1+n} \frac{d t}{(t)^{1 / n}} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left(\frac{t^{1-\frac{1}{n}}}{1-\frac{1}{n}}\right)_1^{1+n} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n(n-1)}\left[(1+n)^{1-\frac{1}{n}}-1\right]\end{aligned}$ <br/><br/>Put $n=\frac{1}{h}$ <br/><br/>When, $n \rightarrow \infty$, then $h \rightarrow 0$ <br/><br/>$$ \begin{aligned} & =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[\left(1+\frac{1}{h}\right)^{1-h}-1\right] \\\\ & =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[1-(1-h)\left(1+\frac{1}{h}\right)+\ldots-1\right] \\\\ & =\lim _{h \rightarrow 0} \frac{h^2}{1-h}\left[-(1-h)\left(1+\frac{1}{h}\right)+\ldots\right] \\\\ & =\lim _{h \rightarrow 0}-h[h+1+\ldots . .]=0 \end{aligned} $$