If f(x)=lim⁡n→∞1(3πtan⁡−12x)2n+5f(x) = \lim_{n \to \infty} \frac{1}{\left( \frac{3}{\pi} \tan^{-1} 2x \right)^{2n} + 5}…

Problem

If f(x)=lim⁡n→∞1(3πtan⁡−12x)2n+5f(x) = \lim_{n \to \infty} \frac{1}{\left( \frac{3}{\pi} \tan^{-1} 2x \right)^{2n} + 5} , then the set of values of xx for which f(x)=0f(x) = 0 , is a. ∣2x∣>3|2x| > \sqrt{3} b. ∣2x∣<3|2x| < \sqrt{3} c. ∣2x∣≥3|2x| \ge \sqrt{3} d. ∣2x∣≤3|2x| \le \sqrt{3}
Intuitive explanation

This problem hinges on understanding how limits of the form lim⁡n→∞1A2n+5\lim_{n \to \infty} \frac{1}{A^{2n} + 5} behave depending on the value of AA. The key insight is that A2nA^{2n} either grows without bound, stays constant, or shrinks to zero as n→∞n \to \infty—and only one of these cases makes the denominator blow up enough for f(x)=0f(x) = 0. Figure out which values of the argument 3πtan⁡−12x\frac{3}{\pi} \tan^{-1} 2x force that behavior, and you've found your answer.

Intuitive explanation

This problem hinges on understanding how limits of the form lim⁡n→∞1A2n+5\lim_{n \to \infty} \frac{1}{A^{2n} + 5} behave depending on the value of AA. The key insight is that A2nA^{2n} either grows without bound, stays constant, or shrinks to zero as n→∞n \to \infty—and only one of these cases makes the denominator blow up enough for f(x)=0f(x) = 0. Figure out which values of the argument 3πtan⁡−12x\frac{3}{\pi} \tan^{-1} 2x force that behavior, and you've found your answer.

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