Question
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}2^{\text{n}-1}\sin\Big(\frac{\text{a}}{2^\text{n}}\Big)$
$\lim\limits_{\text{n}\rightarrow\infty}2^{\text{n}-1}\sin\Big(\frac{\text{a}}{2^\text{n}}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2^\text{n}}{2}\sin\Big(\frac{\text{a}}{2^\text{n}}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2^\text{n}}{2}\sin\frac{\text{a}}{2^\text{n}}$
$\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}\rightarrow0, $ let $\frac{1}{\text{n}}=\text{h}$
$=\lim\limits_{{\text{h}}\rightarrow\infty}\frac{2^{\frac1{\text{h}}}}{2}\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}$
$=\lim\limits_{{\text{h}}\rightarrow\infty}\frac{2^{\frac{1}{\text{h}}}}{2}\frac{\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{\text{a}}{2}$
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