Functions Question 242
Question: $ \underset{\alpha \to \beta }{\mathop{\lim }},
[ \frac{{{\sin }^{2}}\alpha -{{\sin }^{2}}\beta }{{{\alpha }^{2}}-{{\beta }^{2}}} ]= $ [MP PET 2001]
Options:
A) 0
B) 1
C) $ \frac{\sin \beta }{\beta } $
D) $ \frac{\sin 2\beta }{2\beta } $
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Answer:
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{lim}},\frac{2\sin ,\alpha \cos \alpha }{2\alpha }=\underset{\alpha \to \beta }{\mathop{lim}},\frac{\sin 2\alpha }{2\alpha }=\frac{\sin 2\beta }{2\beta } $ .
BETA
