Cauchy series in Euler number

Computing the limit separately first we take $\lim_{n\rightarrow\infty} ( 1+\frac{1}{n+1} )^{n+1}$

$=e^(\lim_{n\rightarrow\infty} ( 1+\frac{1}{n+1} -1)(n+1)$ [As it is in the form $1^\infty$ ]

$=e^(\lim_{n\rightarrow\infty} ( \frac{1}{n+1} )(n+1))$

$=e$

Now

$\lim_{n\rightarrow\infty} ( 1+\frac{1}{n} )^n$

$=e^(\lim_{n\rightarrow\infty} ( 1+\frac{1}{n} -1)n$ [As it is in the form $1^\infty$ ]

$=e^(\lim_{n\rightarrow\infty} ( \frac{1}{n} )n)$

$=e$

Now

$\lim_{n\rightarrow\infty} \sqrt{n} \left( \left( 1+\dfrac{1}{n+1} \right)^{n+1} - \left( 1+\dfrac{1}{n} \right)^{n} \right)$

$=\lim_{n\rightarrow\infty} \sqrt{n}*\lim_{n\rightarrow\infty} (( 1+\frac{1}{n+1} )^{n+1}-( 1+\frac{1}{n} )^n)$

$= \lim_{n\rightarrow\infty} \sqrt{n}(e-e)$ [Using computed limits]

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