Let \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{(1+x^{n})^{1/n}}, \mathrm{x} \in \mathrm{R}-\{-1\}, \mathrm{n} \in \mathrm{N}, \mathrm{n}>2\). If \(\mathrm{f}^{\mathrm{n}}(\mathrm{x})=\) (fofof ...... upto n times) ( x ), then \(\operatorname{Lim}_{n \rightarrow \infty} \int_{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x\) is equal to

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Definite Integrals

Let

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

. If

f^{n} (x) =

(fofof ...... upto n times) ( x ), then

Lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f^{n} (x)) d x

is equal to

Answer: 0

By induction,

f^{n} (x) = \frac{x}{( 1 + n x ^{n} ) ^{1/ n}}

. The integral becomes

lim_{n \to \infty} \int_{0}^{1} \frac{x ^{n - 1}}{( 1 + n x ^{n} ) ^{1/ n}} d x

. Substitute

t = 1 + n x^{n}

, so

x^{n - 1} d x = \frac{d t}{n ^{2}}

. The integral transforms to

lim_{n \to \infty} \frac{1}{n ^{2}} \int_{1}^{1 + n} t^{- 1/ n} d t = lim_{n \to \infty} \frac{1}{n ( n - 1 )} [(1 + n)^{(n - 1) / n} - 1]

. Using series expansion with

h = 1/ n

, this limit evaluates to 0.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer