Let f ( x ) = x ( 1 + x n ) 1 n , x- { - 1 } , n, n 2 . If f n ( x ) = ( f o f o f . . . . . upto n times ) ( x ) , then lim n 0 1 x n - 2 ( f n ( x ) ) ? d x is equal to _____ . [2023]

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Solution

Q.
Let $f (x) = \frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}, x \in ℝ - {- 1}, n \in ℕ, n > 2 .$ If $f^{n} (x) = (f o f o f . . . . . upto n times)$ $(x),$ then $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f^{n} (x)) d x$ is equal to _____ . [2023]

Ans.
(0)

We have, $f (x) = \frac{x}{{(1 + x^{n})}^{1 / n}}$

$\Rightarrow f (f (x)) = \frac{x}{{(1 + 2 x^{n})}^{1 / n}}$

$f (f (f (x))) = \frac{x}{{(1 + 3 x^{n})}^{1 / n}}$

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

Now, $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} \cdot (f^{n} (x)) d x = \lim_{n \to \infty} \int_{0}^{1} \frac{x^{n - 1}}{{(1 + n x^{n})}^{1 / n}} d x$

Put $1 + n x^{n} = t \Rightarrow n^{2} x^{n - 1} d x = d t$

When $x = 0, t = 1$ and $x = 1, t = 1 + n$

$∴$ $\lim_{n \to \infty} \cdot \frac{1}{n^{2}} \int_{1}^{1 + n} \frac{d t}{t^{1 / n}} = \lim_{n \to \infty} \frac{1}{n^{2}} {[\frac{t^{1 - 1 / n}}{1 - 1 / n}]}_{1}^{1 + n}$

$= \lim_{n \to \infty} \frac{1}{n (n - 1)} [{(1 + n)}^{1 - 1 / n} - 1]$

$= \lim_{h \to 0} \frac{h^{2}}{1 - h} [{(1 + \frac{1}{h})}^{1 - h} - 1] (Put n = \frac{1}{h}, n \to \infty, h \to 0)$

$= 0$

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