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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Set of all values of xx such that lim⁡n→∞11+(4tan⁡−1(2πx)π)4n\displaystyle\lim_{n\to\infty}\dfrac{1}{1+\left(\dfrac{4\tan^{-1}(2\pi x)}{\pi}\right)^{4n}} is non-zero and finite number when n∈Nn\in\mathbb{N} is
A(0,12π)\left(0,\dfrac{1}{2\pi}\right)
B(−1π,1π)\left(-\dfrac{1}{\pi},\dfrac{1}{\pi}\right)
C[−12π,12π]\left[-\dfrac{1}{2\pi},\dfrac{1}{2\pi}\right]correct
D(−12π,12π)\left(-\dfrac{1}{2\pi},\dfrac{1}{2\pi}\right)
Solution
Step 1: Let A=4tan⁡−1(2πx)πA=\dfrac{4\tan^{-1}(2\pi x)}{\pi}. The limit is 11+A4n\dfrac{1}{1+A^{4n}}. Step 2: As n→∞n\to\infty: - If ∣A∣<1|A|<1: A4n→0A^{4n}\to 0, limit =1=1. - If ∣A∣=1|A|=1: A4n=1A^{4n}=1, limit =12=\dfrac{1}{2}. - If ∣A∣>1|A|>1: A4n→∞A^{4n}\to\infty, limit =0=0. Step 3: For non-zero and finite limit, need ∣A∣≤1|A|\le 1:
∣4tan⁡−1(2πx)π∣≤1  ⇒  ∣tan⁡−1(2πx)∣≤π4\left|\dfrac{4\tan^{-1}(2\pi x)}{\pi}\right|\le 1\;\Rightarrow\;|\tan^{-1}(2\pi x)|\le\dfrac{\pi}{4}
 Step 4: ∣tan⁡−1(2πx)∣≤π4  ⇒  ∣2πx∣≤1  ⇒  ∣x∣≤12π|\tan^{-1}(2\pi x)|\le\dfrac{\pi}{4}\;\Rightarrow\;|2\pi x|\le 1\;\Rightarrow\;|x|\le\dfrac{1}{2\pi}. Step 5: So x∈[−12π,12π]x\in\left[-\dfrac{1}{2\pi},\dfrac{1}{2\pi}\right]. Correct answer: (3) [−12π,12π]\left[-\dfrac{1}{2\pi},\dfrac{1}{2\pi}\right]
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