Question

Find the interval of θ corresponding to:
$\displaystyle \large{ \lim_{n \to \infty} \frac{ \tan^{n} \theta + 2}{2 { \tan}^{n} \theta + 2 } = \frac{1}{2} \: \: \: \: (0 \leqslant \theta < \frac{\pi}{2} )}$
Please show your work too — thanks!​

Answer 1

With $x = \tan^n(\theta)$, we have

$\dfrac{x+2}{2x+2} = \dfrac12\cdot\dfrac{x+2}{x+1} = \dfrac12\cdot \dfrac{x+1+1}{x+1} = \dfrac12 + \dfrac1{x+1}$

so $\theta$ belongs to some interval such that

$\displaystyle\lim_{n\to\infty}\frac1{\tan^n(\theta)+1} = 0$

In order for this limit to be 0, the denominator must blow up as n gets arbitrarily large. This will happen if

$\left|\tan^n(\theta)\right| = |\tan(\theta)|^n \ge 1 \implies |\tan(\theta)| \ge 1 \implies \theta \ge \dfrac\pi4$

(where $|\tan(\theta)|=\tan(\theta)$ since $0\le\theta<\frac\pi2$, over which $\tan(\theta)\ge0$)

So the interval over which the limit is 1/2 is

$\dfrac\pi4 \le \theta < \dfrac\pi2$