A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) 1
D) ?1
Correct Answer: A
Solution :
\[\underset{a\to 0}{\mathop{\lim }}\,\frac{\sin a-\tan a}{{{\sin }^{3}}a}=\underset{a\to 0}{\mathop{\lim }}\,\frac{\cos a-1}{{{\sin }^{2}}a\cos a}=\underset{a\to 0}{\mathop{\lim }}\,\frac{-(1-\cos a)}{(1-{{\cos }^{2}}a)(\cos a)}\] \[=\underset{a\to 0}{\mathop{\lim }}\,\left[ -\frac{1}{(1+\cos a)\cos a} \right]=-\frac{1}{(1+1)1}=\frac{-1}{2}\]\[\]
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