An Integral Limit

Integrating the expression we have got the limit as :

$L=\lim _{ n\rightarrow \infty }{ \quad n((e-1)-({ (1+\frac { 1 }{ n } ) }^{ n }-\frac { n }{ n+1 } )) }$

Simplifying the expression a bit we get :

$L=\lim _{ n\rightarrow \infty }{ \quad n((e-{ (1+\frac { 1 }{ n } ) }^{ n }) } -\frac { n }{ n+1 }$

Put $x=\frac{1}{n}$ to get the limit as :

$\displaystyle L=\lim _{ x\rightarrow 0^{ + } }{ \quad \frac { (e-{ (1+x) }^{ \frac { 1 }{ x } }) }{ x } } -\frac { 1 }{ x+1 }$

$\displaystyle L=(\lim _{ x\rightarrow 0^{ + } }{ \quad \frac { (e\quad -\quad { e }^{ \frac { ln(1+x) }{ x } }) }{ x } }) -1$

Note the given limit is of $\frac{0}{0}$ indeterminate form

Applying L-Hopital's rule we get :

$\displaystyle L=\lim _{ x\rightarrow 0^{ + } }{ \quad \frac { ({ e }^{ \frac { ln(1+x) }{ x } })(ln(1+x)-\frac { x }{ x+1 } ) }{ 1 } } -1$

$\displaystyle L=(\lim _{ x\rightarrow 0^{ + } }{ { e }^{ \frac { ln(1+x) }{ x } }\quad ) } (\lim _{ x\rightarrow 0^{ + } }{ (ln(1+x)-\frac { x }{ 1+x } ) } )-1$

The first limit equals to $e$ and for second limit applying L-Hopital's rule once again we get.

$\displaystyle L=e.\lim _{ x\rightarrow 0^{ + } }{ (\frac { 1 }{ 2{ (1+x) }^{ 2 } } ) } -1$

Finally we get :

$L=\Large \frac { e }{ 2 } -1$