Limited Trigonometry - II

First, we note that

$\begin{array}{c}\ \tan \theta + \tan \theta \sec 2\theta &= \tan \theta (1 + \dfrac{1}{\cos 2\theta}) \\ &= \dfrac{\sin \theta}{\cos \theta} \dfrac{\cos 2\theta+1}{\cos 2\theta} \\ &= \dfrac{\sin \theta (2\cos^2 \theta)}{\cos \theta \cos 2\theta} \\ &= \dfrac{2\sin \theta \cos \theta}{\cos 2\theta} \\ &= \dfrac{\sin 2\theta}{\cos 2\theta} \\ &= \tan 2\theta \end{array}$

Using this identity, we can simplify $f(x) + \tan \frac{x}{2^n}$ like so:

$\begin{array}{c}\ f(x) + \tan \frac{x}{2^n} &= \tan \frac{x}{2} \sec x + \tan \frac{x}{2^2} \sec \frac{x}{2} + \ldots + \tan \frac{x}{2^n} \sec \frac{x}{2^{n-1}} + \tan \frac{x}{2^n} \\ &= \tan \frac{x}{2} \sec x + \tan \frac{x}{2^2} \sec \frac{x}{2} + \ldots + \tan \frac{x}{2^{n-1}} \sec \frac{x}{2^{n-2}} + \tan \frac{x}{2^{n-1}} \\ &= \ldots \\ &= \tan \frac{x}{2} \sec x + \tan \frac{x}{2} \\ &= \tan x \end{array}$

Thus, our limit is

$\lim_{x \to 0} \left( \dfrac{f(x)+\tan \dfrac{x}{2^n}}{x} \right)^{1/x^2} = \lim_{x \to 0} \left( \dfrac{\tan x}{x} \right)^{1/x^2}$

If we substitute $x = \arctan u$ , using the series expansion $\arctan u = u-\frac{u^3}{3}+\frac{u^5}{5}-\ldots$ , we have

$\lim_{x \to 0} \left( \dfrac{\tan x}{x} \right)^{1/x^2} = \lim_{u \to 0} \left( \dfrac{\arctan u}{u} \right)^{-1/x^2} = \lim_{u \to 0} \left( 1-\frac{u^2}{3} \right)^{-1/x^2}$

Since $u \sim \arctan u = x$ as $u$ approaches 0, we are allowed to rewrite our limit as

$\lim_{u \to 0} \left( 1-\frac{u^2}{3} \right)^{-1/x^2} = \lim_{u \to 0} \left( 1-\frac{u^2}{3} \right)^{-1/u^2} = \lim_{v \to -\infty} \left( 1+\frac{1}{3v} \right)^{v} = e^{1/3}$

by making another substitution $v = -\frac{1}{u^2}$ . The limit follows from the fact that $\displaystyle \lim_{v \to \pm \infty} \left( 1+\frac{a}{v} \right)^{v} = e^a$ .

Hence, we have

$\lim_{x \to 0} \left( \dfrac{f(x)+\tan \dfrac{x}{2^n}}{x} \right)^{1/x^2} = \boxed{e^{1/3}}$