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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 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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