Let f(x)=x(1+x^n)^(1)/(n),x R--1,n N,n2.

Question

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

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

$\underset{n	o \infty}{lim}\int _{0}^{1}{x}^{n-2}({f}^{n}(x))dx$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 0

Accepted Answer

$\begin{gathered} f(x)=\frac{x}{{(1+{x}^{n})}^{1/n}} \ 	herefore f(f(x))=\frac{f(x)}{(1+{[f(x{)}^{n}]}^{1/n}} \ =\frac{\frac{x}{{(1+{x}^{n})}^{1/n}}}{{(1+\frac{{x}^{n}}{1+{x}^{n}})}^{1/n}} \ =\frac{x}{{(1+2{x}^{n})}^{1/n}} \ f(f(f(x)))=\frac{x}{{(1+3{x}^{n})}^{1/n}} \ {f}^{n}(x)=\frac{x}{{(1+n{x}^{n})}^{1/n}} \end{gathered}$

Now, $\underset{n	o \infty}{lim}{\int }_{1}^{0}{x}^{n-2}({f}^{n}(x))dx$

$\begin{gathered} =\underset{n	o \infty}{lim}\int _{0}^{1}{x}^{n-2}\frac{x}{{(1+n{x}^{n})}^{1/n}}dx \ =\underset{n	o \infty}{lim}\int _{0}^{1}\frac{{x}^{n-1}}{{(1+n{x}^{n})}^{1/n}}dx \end{gathered}$

Let $1+n{x}^{n}=t$

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

When, $x=0$, then $t=1$

When, $x=1$, then $t=1+n$

$\begin{gathered} =\underset{n	o \infty}{lim}{\int }_{1+n}^{1}\frac{dt}{{n}^{2}(t{)}^{1/n}}=\underset{n	o \infty}{lim}\frac{1}{{n}^{2}}{\int }_{1+n}^{1}\frac{dt}{(t{)}^{1/n}} \ =\underset{n	o \infty}{lim}\frac{1}{{n}^{2}}{(\frac{{t}^{1-\frac{1}{n}}}{1-\frac{1}{n}})}_{1+n}^{1} \ =\underset{n	o \infty}{lim}\frac{1}{n(n-1)}[(1+n{)}^{1-\frac{1}{n}}-1] \end{gathered}$

Put $n=\frac{1}{h}$

When, $n	o \infty$, then $h	o 0$

$\begin{gathered} =\underset{h	o \infty}{lim}\frac{1}{\frac{1}{h}(\frac{1}{h}-1)}[{(1+\frac{1}{h})}^{1-h}-1] \ =\underset{h	o \infty}{lim}\frac{1}{\frac{1}{h}(\frac{1}{h}-1)}[1-(1-h)(1+\frac{1}{h})+\ldots -1] \ =\underset{h	o 0}{lim}\frac{{h}^{2}}{1-h}[-(1-h)(1+\frac{1}{h})+\ldots ] \ =\underset{h	o 0}{lim}-h[h+1+\ldots ..]=0 \end{gathered}$

Solutions · 1