Verified answer

Let y= (1+ 1/n)^(n+2)

Ln(y)= (n+2)ln(1+1/n)

= ln[ 1+ 1/n] ] /[1/(n+2)]

now this fraction approaches 0/0, so we apply l'hopital's

-> [1/(1+ 1/n)]* -n^-2

--------------------------

-(n+2)^-2

= [n/(n+1)] (n+2)^2

------------------------

n^2

= n(n+2)^2

---------------

n^2 * (n+1)

-> 1 as n-> infinity

Since lny -> 1

Then y-> e^1

Y-> e

I hope this helps!