If , then the value of θ is

Engineering

Mathematics

One to The Power Infinity Form

Question

If $\underset{x \to 0}{Lim} {[1 + x \ln (1 + b^{2})]}^{\frac{1}{x}} = 2 {bsin}^{2} θ, b > 0 and θ \in (- π, π]$ , then the value of θ is

$\pm \frac{π}{4}$

$\pm \frac{π}{2}$

$\pm \frac{π}{3}$

$\pm \frac{π}{6}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\lim_{x \to \infty}$ [1 + x ln(1 + b²)]^1/x= 2b sin² θ b > 0; θ∈(− π, π)

$\lim_{x \to \infty} {({[1 + xℓn (1 + b^{2})]}^{\frac{1}{x ℓn (1 + b^{2})}})}^{ℓn (1 + b^{2})} = 2 b \sin^{2} θ$

eln(1 + b²) = 2b sin² θ

1 + b² = 2b sin² θ

$2 \sin^{2} θ = b + \frac{1}{b}$

$RHS = b + \frac{1}{b} \geq 2 as b > 0$

But LHS = 2 sin² θ ≤ 2

Only possibility

2 sin² θ = 2

Sin²θ = 1

$θ = \pm \frac{π}{2}$