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Integral Calculus: JEE Main Mathematics Question with Solution

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

Yes! Since , we have , so , immediately giving the limit as 0.

Let

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

. If

f^n(x)=(

fofof

\ldots .

. 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

\text{\quad\quad\quad}

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Step-by-step solutionView

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 $x \in \mathbb{R} \setminus \{-1\}$ , $n \in \mathbb{N}, n > 2$ , 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: $f(f(x)) = \frac{f(x)}{(1 + (f(x))^n)^{1/n}} = \frac{\frac{x}{(1+x^n)^{1/n}}}{\left(1 + \frac{x^n}{1+x^n}\right)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$ . By induction, $f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$ .

04execute

Substitute $f^n(x)$ into the integral: $I_n = \int_0^1 x^{n-2} f^n(x) \, dx = \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}} \, dx$ . Let $t = 1 + nx^n$ , so $dt = n^2 x^{n-1} dx \implies x^{n-1} dx = \frac{dt}{n^2}$ . When $x = 0$ , $t = 1$ ; when $x = 1$ , $t = 1 + n$ . Thus, $I_n = \frac{1}{n^2} \int_1^{1+n} t^{-1/n} \, dt = \frac{1}{n^2} \left[ \frac{t^{1 - 1/n}}{1 - 1/n} \right]_1^{1+n} = \frac{1}{n(n-1)} \left[ (1+n)^{\frac{n-1}{n}} - 1 \right]$ .

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 $n(n-1) \sim n^2$ . $\lim_{n \to \infty} \frac{(1+n)^{\frac{n-1}{n}} - 1}{n(n-1)} = \lim_{n \to \infty} \frac{O(n)}{n^2} = 0$ .

06verify

Bounding the integrand on $[0,1]$ : $0 \le \frac{x^{n-1}}{(1+nx^n)^{1/n}} \le x^{n-1}$ . Thus $0 \le I_n \le \int_0^1 x^{n-1} dx = \frac{1}{n} \to 0$ as $n \to \infty$ . By Squeeze Theorem, the limit is strictly 0.

07given

$f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$ and the integral is $I_n = \int_0^1 x^{n-2} f^n(x) dx = \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}} dx$ .

08approach

For $x \in [0, 1]$ , $1+nx^n \ge 1$ , which gives a sharp upper bound on the integrand.

09execute

Since $(1+nx^n)^{1/n} \ge 1$ for all $x \in [0,1]$ : $0 \le \frac{x^{n-1}}{(1+nx^n)^{1/n}} \le x^{n-1}$ Integrating over $[0,1]$ : $0 \le I_n \le \int_0^1 x^{n-1} dx = \frac{1}{n}$ 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.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

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

Yes! Since $1+nx^n \ge 1$ , we have $0 \le f^n(x) \le x$ , so $0 \le \int_0^1 x^{n-1} dx = 1/n \to 0$ , 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 $x \in \mathbb{R} \setminus \{-1\}$ , $n \in \mathbb{N}, n > 2$ , 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: $f(f(x)) = \frac{f(x)}{(1 + (f(x))^n)^{1/n}} = \frac{\frac{x}{(1+x^n)^{1/n}}}{\left(1 + \frac{x^n}{1+x^n}\right)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$ . By induction, $f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$ .
execute: Substitute $f^n(x)$ into the integral: $I_n = \int_0^1 x^{n-2} f^n(x) \, dx = \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}} \, dx$ . Let $t = 1 + nx^n$ , so $dt = n^2 x^{n-1} dx \implies x^{n-1} dx = \frac{dt}{n^2}$ . When $x = 0$ , $t = 1$ ; when $x = 1$ , $t = 1 + n$ . Thus, $I_n = \frac{1}{n^2} \int_1^{1+n} t^{-1/n} \, dt = \frac{1}{n^2} \left[ \frac{t^{1 - 1/n}}{1 - 1/n} \right]_1^{1+n} = \frac{1}{n(n-1)} \left[ (1+n)^{\frac{n-1}{n}} - 1 \right]$ .
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 $t = 1 + n$ . $\lim_{n \to \infty} \frac{(1+n)^{\frac{n-1}{n}} - 1}{n(n-1)} = \lim_{n \to \infty} \frac{O(n)}{n^2} = 0$ .
verify: Bounding the integrand on $[0,1]$ : $0 \le \frac{x^{n-1}}{(1+nx^n)^{1/n}} \le x^{n-1}$ . Thus $0 \le I_n \le \int_0^1 x^{n-1} dx = \frac{1}{n} \to 0$ as $n \to \infty$ . By Squeeze Theorem, the limit is strictly 0.
given: $f^n(x) = \frac{x}{(1+nx^n)^{1/n}}$ and the integral is $I_n = \int_0^1 x^{n-2} f^n(x) dx = \int_0^1 \frac{x^{n-1}}{(1+nx^n)^{1/n}} dx$ .
approach: For $x \in [0, 1]$ , $1+nx^n \ge 1$ , which gives a sharp upper bound on the integrand.
execute: Since $(1+nx^n)^{1/n} \ge 1$ for all $x \in [0,1]$ : $0 \le \frac{x^{n-1}}{(1+nx^n)^{1/n}} \le x^{n-1}$ Integrating over $[0,1]$ : $0 \le I_n \le \int_0^1 x^{n-1} dx = \frac{1}{n}$ 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.

Integral Calculus: JEE Main Mathematics Question with Solution

The limit of the integral as n→∞n \to \inftyn→∞ is equal to 0.

01given

02approach

03execute

04execute

05execute

06verify

07given

08approach

09execute

✓verify

Students also ask

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