Limits up to the Limit -1!

$\begin{aligned} f(x) & = \lim_{n \to \infty} \left( \tan \frac x2 \sec x + \tan \frac x{2^2} \sec \frac x2 + \tan \frac x{2^3} \sec \frac x{2^2} + \cdots + \tan \frac x{2^n} \sec \frac x{2^{n-1}} \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n \tan \frac x{2^k} \sec \frac x{2^{n-1}} & \small \color{#3D99F6} \text{Let }t = \tan \frac x{2^k} \\ & = \lim_{n \to \infty} \sum_{k=1}^n t \cdot \frac {1+t^2}{1-t^2} = \lim_{n \to \infty} \sum_{k=1}^n t \cdot \frac {2-(1-t^2)}{1-t^2} = \lim_{n \to \infty} \sum_{k=1}^n \left( \frac {2t}{1-t^2} - t\right) & \small \color{#3D99F6} \text{Then }\cos \frac x{2^{k-1}} = \frac {1-t^2}{1+t^2} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \left(\tan \frac x{2^{k-1}} - \tan \frac x{2^k} \right) = \lim_{n \to \infty} \left(\tan x - \tan \frac 1{2^n}\right) \\ & = \tan x \end{aligned}$

Therefore, $f\left(\frac \pi 4\right) = \tan \frac \pi 4 = \boxed 1$ .