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Lim N → ∞ 2 N − 1 Sin ( a 2 N )

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प्रश्न

\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\] 

 

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उत्तर

\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[ = \lim_{n \to \infty} \frac{2^n}{2} \times \frac{\sin \left( \frac{a}{2^n} \right)}{\left( \frac{a}{2^n} \right)} \times \left( \frac{a}{2^n} \right)\]
\[Let y = \frac{a}{2^n}\]
\[If n \to \infty , then   y \to 0 . \]
\[ = \lim_{y \to 0} \frac{a}{2} \times \left( \frac{\sin y}{y} \right)\]
\[ \Rightarrow \frac{a}{2}\] 

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 25 | पृष्ठ ६२

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