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Limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
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Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
$\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}=\text{h} \rightarrow0$
$=\frac{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\Bigg)}{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\Bigg)}$
$=\frac{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}$
$=\frac{1\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{1\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}=\frac{\text{a}}{\text{b}}$

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