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Question

If $$y\left( n \right) ={ e }^{ x }{ e }^{ { x }^{ 2 } }...{ e }^{ { x }^{ n } },0<x<1$$ then $$\displaystyle \lim _{ n\rightarrow \infty  }{ \frac { d\ y\left( n \right)  }{ dx }  } $$ at $$\displaystyle\frac{1}{2}$$ is


A
e
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B
4e
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C
2e
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D
3e
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Solution

The correct option is D $$4e$$

$$\displaystyle y\left( n \right) ={ e }^{ x+{ x }^{ 2 }+...+{ x }^{ n } }={ e }^{ \dfrac { x\left( 1-{ x }^{ n } \right)  }{ 1-x }  }$$


So $$\displaystyle \dfrac { dy\left( n \right)  }{ dx } ={ e }^{ \dfrac { x\left( 1-{ x }^{ n } \right)  }{ 1-x }  }\times \dfrac { d }{ dx } \left( \dfrac { x\left( 1-{ x }^{ n } \right)  }{ 1-x }  \right) $$


$$\displaystyle \lim _{ n\rightarrow \infty  }{ \dfrac { dy\left( n \right)  }{ dx } = } \lim _{ n\rightarrow \infty  }{ { e }^{ \dfrac { x }{ 1-x } -\dfrac { { x }^{ n+1 } }{ 1-x }  } } \dfrac { d }{ dx } \left[ \lim _{ n\rightarrow \infty  }{ \dfrac { x }{ 1-x } -\dfrac { { x }^{ n+1 } }{ 1-x }  }  \right] $$


$$\displaystyle ={ e }^{ \dfrac { x }{ dx }  }\times \dfrac { d }{ dx } \left( \dfrac { x }{ dx }  \right) ={ e }^{ \dfrac { x }{ 1-x }  }\dfrac { d }{ dx } \left( -1+\dfrac { 1 }{ 1-x }  \right) $$


$$\displaystyle ={ e }^{ \dfrac { x }{ 1-x }  }.\dfrac { d }{ dx } \dfrac { 1 }{ { \left( 1-x \right)  }^{ 2 } } $$


$$\because $$ for $$\displaystyle x\in \left( 0,1 \right) ;{ x }^{ n }\rightarrow 0$$ 


$$\displaystyle \therefore \lim _{ n\rightarrow \infty  }{ \dfrac { { x }^{ n+1 } }{ 1-x }  } =0$$


$$\displaystyle \Rightarrow \lim _{ n\rightarrow \infty  }{ \left( \dfrac { dy\left( n \right)  }{ dx }  \right) _{ x=\dfrac { 1 }{ 2 }  }=4e } $$

 


Mathematics

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