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$\lim _{\alpha \rightarrow \beta}\left[\frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}\right]=$
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Verified Answer
The correct answer is:
$\frac{\sin 2 \beta}{2 \beta}$
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}$
Applying L-Hospital's rule,
$\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha}{2 \alpha}=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha}=\frac{\sin 2 \beta}{2 \beta}$
Applying L-Hospital's rule,
$\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha}{2 \alpha}=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha}=\frac{\sin 2 \beta}{2 \beta}$
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