Question

$\underset{n\to \infty }{\lim }\frac{\sin \left({\displaystyle \frac{a}{2^n}}\right)}{\sin \left({\displaystyle \frac{b}{2^n}}\right)}$

Answer 1

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{\sin \left({\displaystyle \frac{a}{2^n}}\right)}{\sin \left({\displaystyle \frac{b}{2^n}}\right)} \\ =\underset{n\to \infty }{\lim }\frac{\left({\displaystyle \frac{a}{2^n}}\right)\sin {\displaystyle \left(\frac{a}{2^n}\right)}}{\left(\frac{a}{2^n}\right)\times \left(\frac{b}{2^n}\right)\times \left(\frac{\sin \left({\displaystyle \frac{b}{2^n}}\right)}{{\displaystyle \frac{b}{2^n}}}\right)} \\ \mathrm{Let}: \\ y=\frac{a}{2^n} \\ z=\frac{b}{2^n} \\ \mathrm{If}n\to \infty ,\mathrm{then}y\to 0\mathrm{and}z\to 0. \\ =\frac{y}{z}\underset{y\to 0}{\lim }\frac{\sin y}{y}\times \frac{1}{\underset{z\to 0}{\lim }{\displaystyle \frac{\sin z}{z}}} \\ =\frac{y}{z}\times 1\times \frac{1}{1} \\ =\frac{{\displaystyle \frac{a}{2^n}\times 1}}{{\displaystyle \frac{b}{2^n}\times 1}} \\ =\frac{a}{b} \end{gathered}$$