A) \[\frac{1}{3}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{12}\]
Correct Answer: D
Solution :
[d]: Put cos x=t \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(cos\,x)}^{1/3}}-{{(cos\,x)}^{1/2}}}{{{\sin }^{2}}x}=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{\frac{1}{3}}}-{{t}^{\frac{1}{2}}}}{1-{{t}^{2}}}\] \[=\underset{t\to 1}{\mathop{\lim }}\,\frac{\frac{1}{3}{{t}^{\frac{-2}{3}}}-\frac{1}{2}{{t}^{\frac{-1}{2}}}}{-2t}=\frac{\frac{1}{3}-\frac{1}{2}}{-2}=\frac{1}{12}\](By L.H. Rule)
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