JEE Main 2023 (Online) 6th April Evening ShiftQuestion 1
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 $f^{n}(x)=(f\circ f\circ f\ldots\text{ upto }n\text{ times})(x)$, then $\lim_{n\to\infty}\int_{0}^{1}x^{n-2}\left(f^{n}(x)\right)dx$ is equal to $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$0$
Solution
Concept. Find a pattern for repeated composition, then compute the integral exactly. Formulas used. (1) $f^k(x)=\frac{x}{(1+kx^n)^{1/n}}$(2) $\int(1+nu)^{-1/n}du=\frac{(1+nu)^{1-1/n}}{n-1}$Step 1 - Two steps. $f(x)^n=\frac{x^n}{1+x^n}$, so $1+f(x)^n=\frac{1+2x^n}{1+x^n}$ and $$f(f(x))=\frac{f(x)}{(1+f^n)^{1/n}}=\frac{x}{(1+2x^n)^{1/n}}$$ Repeating gives (1); so $f^n(x)=\frac{x}{(1+nx^n)^{1/n}}$. Step 2 - Integral. $I_n=\int_0^1\frac{x^{n-1}}{(1+nx^n)^{1/n}}dx$. Put $u=x^n$, $du=nx^{n-1}dx$: $$I_n=\frac1n\int_0^1(1+nu)^{-1/n}du=\frac{(n+1)^{1-1/n}-1}{n(n-1)}$$ using (2). Step 3 - Limit. $(n+1)^{1/n}\to1$, so $(n+1)^{1-1/n}\sim n$ and $I_n\sim\frac{n}{n^2}\to0$. Answer: 0