What is $\lim_{n \to \infty} 2^n \tan\left(\frac{a}{2^n}\right)$? $$\lim_{n\to \infty}{2^n\cdot \tan\left(\frac{a}{2^n}\right)}$$

Experimenting with Wolfram Alpha, I came to suspect the limit is ${a}$. Any help on this matter? I couldn't find it with with half-angle formulas and the series expansion is not very productive either. A similar  result is used (without proof) in the answer of Rohan Shindes question:  Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$ yet eventhough I invested some effort, I was unable to find the solution. Any help would be appreciated.
Write your term in the form $$\frac{\tan\left(\frac{a}{2^n}\right)}{\frac{a}{2^n}}\times a$$
For $a=0$ we obtain:
$$\lim_{n\rightarrow+\infty}2^n\tan\frac{a}{2^n}=0.$$
For $a\neq0$ we obtain:
$$\lim_{n\rightarrow+\infty}2^n\tan\frac{a}{2^n}=a\lim_{n\rightarrow+\infty}\frac{\tan\frac{a}{2^n}}{\frac{a}{2^n}}=a,$$ which says that for all real $a$ we have:
$$\lim_{n\rightarrow+\infty}2^n\tan\frac{a}{2^n}=a.$$