Question

Find the interval of θ corresponding to:


`\displaystyle \large{ \lim_(n \to \infty) \frac{ \tan^(n) \theta + 2}{2 { \tan}^(n) \theta + 2 } = (1)/(2) \: \: \: \: (0 \leqslant \theta < (\pi)/(2) )}`
Please show your work too — thanks!​

Answer 1

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

`(x+2)/(2x+2) = \frac12\cdot(x+2)/(x+1) = \frac12\cdot (x+1+1)/(x+1) = \frac12 + \frac1{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 \frac\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

`\frac\pi4 \le \theta < \frac\pi2`