\( \lim_{\alpha \to 0} \frac{1 - \sin^2 \alpha - \cos \alpha}{\alpha^2} \) এর মান কোনটি?

Explanation: Hints: La' Hospital Rule: \(\lim_{\alpha \to 0} \frac{1 - \sin^2\alpha - \cos\alpha}{\alpha^2}\) Solve: \(\lim_{\alpha \to 0} \frac{0 - 2\sin\alpha\cos\alpha + \sin\alpha}{2\alpha} \implies \lim_{\alpha \to 0} \frac{-\sin 2\alpha + \sin \alpha}{2\alpha}\) [Using La' Hospital again] \(\lim_{\alpha \to 0} \frac{-2\cos 2\alpha + \cos \alpha}{2}\) \(\implies \frac{-2\cos 2\cdot 0 + \cos 0}{2} \implies \frac{-2 + 1}{2} = -\frac{1}{2}\) Ans. (A)

Another Explanation (5): ```html

প্রশ্ন: \( \lim_{\alpha \to 0} \frac{1 - \sin^2 \alpha - \cos \alpha}{\alpha^2} \) এর মান নির্ণয় করো। 🤔

সমাধান:

আমরা জানি, \( \sin^2 \alpha + \cos^2 \alpha = 1 \). সুতরাং, \( 1 - \sin^2 \alpha = \cos^2 \alpha \). 🤓

অতএব, \( \lim_{\alpha \to 0} \frac{1 - \sin^2 \alpha - \cos \alpha}{\alpha^2} = \lim_{\alpha \to 0} \frac{\cos^2 \alpha - \cos \alpha}{\alpha^2} \) । 😮

\( = \lim_{\alpha \to 0} \frac{\cos \alpha (\cos \alpha - 1)}{\alpha^2} \) । 🤗

আমরা জানি, \( \lim_{\alpha \to 0} \cos \alpha = 1 \). 👍

এখন, \( \cos \alpha - 1 = -2 \sin^2 \frac{\alpha}{2} \) । 😎

সুতরাং, \( \lim_{\alpha \to 0} \frac{\cos \alpha (\cos \alpha - 1)}{\alpha^2} = \lim_{\alpha \to 0} \frac{\cos \alpha (-2 \sin^2 \frac{\alpha}{2})}{\alpha^2} \) । ✨

\( = \lim_{\alpha \to 0} \cos \alpha \cdot \lim_{\alpha \to 0} \frac{-2 \sin^2 \frac{\alpha}{2}}{\alpha^2} \) । 🥳

\( = 1 \cdot (-2) \lim_{\alpha \to 0} \frac{\sin^2 \frac{\alpha}{2}}{\alpha^2} \) । 🤩

\( = -2 \lim_{\alpha \to 0} \frac{\sin^2 \frac{\alpha}{2}}{4 \cdot \frac{\alpha^2}{4}} \) । 🤯

\( = -2 \cdot \frac{1}{4} \lim_{\alpha \to 0} \frac{\sin^2 \frac{\alpha}{2}}{\frac{\alpha^2}{4}} \) । 😴

\( = -\frac{1}{2} \lim_{\alpha \to 0} \left( \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2}} \right)^2 \) । 🥰

আমরা জানি, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). 😴

সুতরাং, \( -\frac{1}{2} \cdot (1)^2 = -\frac{1}{2} = -0.5 \) । ✅

অতএব, \( \lim_{\alpha \to 0} \frac{1 - \sin^2 \alpha - \cos \alpha}{\alpha^2} = -0.5 \) ।

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