Question

$\underset{n\to \infty }{lim}\frac{\sin \left(\frac{a}{2^n}\right)}{\sin \left(\frac{b}{2^{n+1}}\right)}=$

• A. $\frac{a}{2b}$
• B. $\frac{2a}{b}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{2a}{b}$

$\underset{n\to \infty }{lim}\frac{\sin (\frac{a}{2^n})}{\sin (\frac{b}{2^{n+1}})}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{\frac{\sin \left(\frac{a}{2^n}\right)}{\left(\frac{a}{2^n}\right)}\frac{a}{2^n}}{\frac{\sin \left(\frac{b}{2^{n+1}}\right)}{\frac{b}{2^{n+1}}}\frac{b}{2^{n+1}}}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{\frac{a}{2^n}}{\frac{b}{2^{n+1}}}$
$\Rightarrow \frac{2a}{b}$
Hence, option 'B' is correct.