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Question

$$\displaystyle\lim _{ x\rightarrow \pi /2 }{ \cfrac { \left( 1-\tan { x/2 }  \right) \left( 1-\sin { x }  \right)  }{ (1-\tan { x/2 } ){ \left( \pi -2x \right)  }^{ 3 } }  } $$ is equal to


A
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B
1/8
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C
0
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D
1/32
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Solution

The correct option is C $$1/32$$
We have
$$\lim _{ x\rightarrow \pi /2 }{ \left( \cfrac { 1-\tan { x/2 }  }{ 1+\tan { x/2 }  }  \right) \left\{ \cfrac { 1-\sin { x }  }{ { \left( \pi -2x \right)  }^{ 3 } }  \right\}  } $$
$$=\cfrac { 1 }{ 64 }\displaystyle \lim _{ x\rightarrow \pi /2 }{ \cfrac { \tan { \left( \cfrac { \pi  }{ 4 } -\cfrac { x }{ 2 }  \right)  }  }{ \left( \cfrac { \pi  }{ 4 } -\cfrac { x }{ 2 }  \right)  }  } \cfrac { \left[ 1-\cos { \left( \pi /2-x \right)  }  \right]  }{ { \left( \cfrac { \pi  }{ 4 } -\cfrac { x }{ 2 }  \right)  }^{ 2 } }  $$
$$=1/64\times 1\times 2=1/32$$

Mathematics

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