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(D) We need to evaluate the limit: $\mathop {\lim }\limits_{\alpha 	o \beta } \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2

Vedclass · Accepted Answer

$\frac{{\sin 2\beta }}{{2\beta }}$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{\alpha 	o \beta } \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}$Since the limit is in the indeterminate form $\frac{0}{0}$ as $\alpha 	o \beta$,we apply $L'	ext{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $\alpha$:$\frac{d}{d\alpha} (\sin^2 \alpha - \sin^2 \beta) = 2 \sin \alpha \cos \alpha = \sin 2\alpha$$\frac{d}{d\alpha} (\alpha^2 - \beta^2) = 2\alpha$Now,taking the limit as $\alpha 	o \beta$:$\mathop {\lim }\limits_{\alpha 	o \beta } \frac{{\sin 2\alpha }}{{2\alpha }} = \frac{{\sin 2\beta }}{{2\beta }}$.

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $

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