JEE MainMathematics

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Limit of an Integral Involving Iterated Composite Rational Functions

Q: Could we have avoided computing the integral altogether using the Squeeze Theorem?

Yes! Since 1+nx^n ≥ 1, we have 0 ≤ f^n(x) ≤ x, so 0 ≤ integral from 0 to 1 of x^(n-1) dx = 1/n → 0, immediately giving the limit as 0.

Let f(x)=x/((1+x^n)^(1/n)), x ∈ R-{-1}, n ∈ N, n>2. If f^n(x)=(fofof ….. upto n times)(x), then Lim_(n → ∞) integral from 0 to 1 of x^(n-2)(f^n(x)) d x is equal to

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Correct answer

The limit of the integral as $n \to \infty$ is equal to 0.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Given $f (x) = \frac{x}{(1 + x^{n})^{1 / n}}$ for $x \in R ∖ {- 1}$ , $n \in N, n > 2$ , and $f^{n} (x) = (f \circ f \circ \dots \circ f) (x)$ ( $n$ times).

02Approach

Compute the general form for the $k$ -th composition $f^{k} (x)$ by induction, substitute $f^{n} (x)$ into the integral, use the substitution $t = 1 + n x^{n}$ , and evaluate the limit as $n \to \infty$ .

03Execute

Compute repeated compositions: $f (f (x)) = \frac{f (x)}{(1 + (f (x))^{n})^{1 / n}} = \frac{\frac{x}{(1 + x^{n})^{1 / n}}}{{(1 + \frac{x^{n}}{1 + x^{n}})}^{1 / n}} = \frac{x}{(1 + 2 x^{n})^{1 / n}}$ . By induction, $f^{n} (x) = \frac{x}{(1 + n x^{n})^{1 / n}}$ .

04Execute

Substitute $f^{n} (x)$ into the integral: $I_{n} = \int_{0}^{1} x^{n - 2} f^{n} (x) d x = \int_{0}^{1} \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} d x$ . Let $t = 1 + n x^{n}$ , so $d t = n^{2} x^{n - 1} d x ⟹ x^{n - 1} d x = \frac{d t}{n^{2}}$ . When $x = 0$ , $t = 1$ ; when $x = 1$ , $t = 1 + n$ . Thus, $I_{n} = \frac{1}{n^{2}} \int_{1}^{1 + n} t^{- 1 / n} d t = \frac{1}{n^{2}} {[\frac{t^{1 - 1 / n}}{1 - 1 / n}]}_{1}^{1 + n} = \frac{1}{n (n - 1)} [(1 + n)^{\frac{n - 1}{n}} - 1]$ .

05Execute

Evaluate $\lim_{n \to \infty} I_{n}$ : As $n \to \infty$ , $(1 + n)^{\frac{n - 1}{n}} = (1 + n)^{1 - 1 / n} = \frac{1 + n}{(1 + n)^{1 / n}} \approx \frac{n}{1} = n$ . Therefore, the numerator behaves as $n$ , while the denominator is $n (n - 1) \sim n^{2}$ . $\lim_{n \to \infty} \frac{(1 + n)^{\frac{n - 1}{n}} - 1}{n (n - 1)} = \lim_{n \to \infty} \frac{O (n)}{n^{2}} = 0$ .

✓Verify

Bounding the integrand on $[0, 1]$ : $0 \leq \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} \leq x^{n - 1}$ . Thus $0 \leq I_{n} \leq \int_{0}^{1} x^{n - 1} d x = \frac{1}{n} \to 0$ as $n \to \infty$ . By Squeeze Theorem, the limit is strictly 0.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

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Students also ask

Could we have avoided computing the integral altogether using the Squeeze Theorem?

Yes! Since $1 + n x^{n} \geq 1$ , we have $0 \leq f^{n} (x) \leq x$ , so $0 \leq \int_{0}^{1} x^{n - 1} d x = 1 / n \to 0$ , immediately giving the limit as 0.

Answer

The limit of the integral as $n \to \infty$ is equal to 0.

Why each option works or fails

Step-by-step solution

Given: Given $f (x) = \frac{x}{(1 + x^{n})^{1 / n}}$ for $x \in R ∖ {- 1}$ , $n \in N, n > 2$ , and $f^{n} (x) = (f \circ f \circ \dots \circ f) (x)$ ( $n$ times).
Approach: Compute the general form for the $k$ -th composition $f^{k} (x)$ by induction, substitute $f^{n} (x)$ into the integral, use the substitution $t = 1 + n x^{n}$ , and evaluate the limit as $n \to \infty$ .
Execute: Compute repeated compositions: $f (f (x)) = \frac{f (x)}{(1 + (f (x))^{n})^{1 / n}} = \frac{\frac{x}{(1 + x^{n})^{1 / n}}}{{(1 + \frac{x^{n}}{1 + x^{n}})}^{1 / n}} = \frac{x}{(1 + 2 x^{n})^{1 / n}}$ . By induction, $f^{n} (x) = \frac{x}{(1 + n x^{n})^{1 / n}}$ .
Execute: Substitute $f^{n} (x)$ into the integral: $I_{n} = \int_{0}^{1} x^{n - 2} f^{n} (x) d x = \int_{0}^{1} \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} d x$ . Let $t = 1 + n x^{n}$ , so $d t = n^{2} x^{n - 1} d x ⟹ x^{n - 1} d x = \frac{d t}{n^{2}}$ . When $x = 0$ , $t = 1$ ; when $x = 1$ , $t = 1 + n$ . Thus, $I_{n} = \frac{1}{n^{2}} \int_{1}^{1 + n} t^{- 1 / n} d t = \frac{1}{n^{2}} {[\frac{t^{1 - 1 / n}}{1 - 1 / n}]}_{1}^{1 + n} = \frac{1}{n (n - 1)} [(1 + n)^{\frac{n - 1}{n}} - 1]$ .
Execute: Evaluate $\lim_{n \to \infty} I_{n}$ : As $n \to \infty$ , $(1 + n)^{\frac{n - 1}{n}} = (1 + n)^{1 - 1 / n} = \frac{1 + n}{(1 + n)^{1 / n}} \approx \frac{n}{1} = n$ . Therefore, the numerator behaves as $n$ , while the denominator is $n (n - 1) \sim n^{2}$ . $\lim_{n \to \infty} \frac{(1 + n)^{\frac{n - 1}{n}} - 1}{n (n - 1)} = \lim_{n \to \infty} \frac{O (n)}{n^{2}} = 0$ .
Verify: Bounding the integrand on $[0, 1]$ : $0 \leq \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} \leq x^{n - 1}$ . Thus $0 \leq I_{n} \leq \int_{0}^{1} x^{n - 1} d x = \frac{1}{n} \to 0$ as $n \to \infty$ . By Squeeze Theorem, the limit is strictly 0.

Shortcut:

When to use it: Use bounding / squeeze theorem whenever an integral with powers of x on [0,1] is taken in a limit n -> infinity.

Given: $f^{n} (x) = \frac{x}{(1 + n x^{n})^{1 / n}}$ and the integral is $I_{n} = \int_{0}^{1} x^{n - 2} f^{n} (x) d x = \int_{0}^{1} \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} d x$ .

Approach: For $x \in [0, 1]$ , $1 + n x^{n} \geq 1$ , which gives a sharp upper bound on the integrand.

Execute: Since $(1 + n x^{n})^{1 / n} \geq 1$ for all $x \in [0, 1]$ : $0 \leq \frac{x^{n - 1}}{(1 + n x^{n})^{1 / n}} \leq x^{n - 1}$ Integrating over $[0, 1]$ : $0 \leq I_{n} \leq \int_{0}^{1} x^{n - 1} d x = \frac{1}{n}$ Taking the limit as $n \to \infty$ , $\lim_{n \to \infty} \frac{1}{n} = 0$ , so $\lim_{n \to \infty} I_{n} = 0$ .

Verify: The integrand is non-negative and bounded above by a function whose integral vanishes as $n \to \infty$ , ensuring the result is precisely 0.

Limit of an Integral Involving Iterated Composite Rational Functions

What feels right?

The limit of the integral as $n \to \infty$ is equal to 0.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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