Suppose the limit L = lim n → ∞ n ∫ 0 1 1 1 + x 2 n d x exists and is larger than 1 2 , then

We have, L = lim n → ∞ n ∫ 0 1 d x 1 + x 2 n We know that, 1 + x 2 n > 1 + n x 2 ⇒ 1 1 + x 2 n < 1 1 + n x 2 ⇒ ∫ 0 1 d x 1 + x 2 n < ∫ 0 1 1 1 + n x 2 d x ⇒ ∫ 0 1 1 1 + x 2 n < 1 n tan - 1 n x 0 1 = 1 n tan - 1 n L = lim n → ∞ n ∫ 0 1 1 1 + x 2 n d x < lim n → ∞ n 1 n tan - 1 n ∴ L = π 2

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Medium2 minPYQ_2019

MathematicsMediumsingle choice

Suppose the limit $L = \lim_{n \to \infty} \sqrt{n} \int_{0}^{1} \frac{1}{{(1 + x^{2})}^{n}} d x$ exists and is larger than $\frac{1}{2}$ , then

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Answer:

Solution:

We have,

$L = \lim_{n \to \infty} \sqrt{n} \int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{n}}$

We know that, ${(1 + x^{2})}^{n} > 1 + n x^{2}$

$\Rightarrow \frac{1}{{(1 + x^{2})}^{n}} < \frac{1}{1 + n x^{2}}$

$\Rightarrow \int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{n}} < \int_{0}^{1} \frac{1}{1 + n x^{2}} d x$

$\Rightarrow \int_{0}^{1} \frac{1}{{(1 + x^{2})}^{n}} < \frac{1}{\sqrt{n}} {[\tan^{- 1} \sqrt{n} x]}_{0}^{1}$

$= \frac{1}{\sqrt{n}} \tan^{- 1} \sqrt{n}$

$L = \lim_{n \to \infty} \sqrt{n} \int_{0}^{1} \frac{1}{{(1 + x^{2})}^{n}} d x < \lim_{n \to \infty} \sqrt{n} \frac{1}{\sqrt{n}} \tan^{- 1} \sqrt{n}$

$∴ L = \frac{π}{2}$

Stream:BITSATSubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2019

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