A) 0
B) 1
C) \[\frac{\sin \beta }{\beta }\]
D) \[\frac{\sin 2\beta }{2\beta }\]
Correct Answer: D
Solution :
\[\underset{\alpha \to \beta }{\mathop{\lim }}\,\frac{{{\sin }^{2}}\alpha -{{\sin }^{2}}\beta }{{{\alpha }^{2}}-{{\beta }^{2}}}\] Applying L-Hospital?s rule, \[\underset{\alpha \to \beta }{\mathop{\text{lim}}}\,\frac{2\sin \,\alpha \,\,\cos \alpha }{2\alpha }=\underset{\alpha \to \beta }{\mathop{\text{lim}}}\,\frac{\sin \,\,2\alpha }{2\alpha }=\frac{\sin \,\,2\beta }{2\beta }\].
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