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
B
4e
C
2e
D
3e

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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