Verified answer

actually L'H is not needed IF you know that [ 1 + p / w ]^w---> e^p as w---> ∞

let w = x - a , w---> ∞ as x ---> ∞...[ ( x + a) / ( x -a ) ]^x = [ ( w + 2a) / w }^(w + a)

= [ 1 + 2a / w ]^w [ 1 + 2a / w}^a ---> e^(2a) { 1 } as w---> ∞

using L'H you need to rewrite as e^{ [ ln(x+a) - ln(x-a) ] / (1/x)}---> e^(0/0)

now use L'H on the ( 0 / 0 ).

y = [(x+a)/(x-a)]^x = [x/(x-a) + a/(x-a)]^x

lny = xln[(x+a)/(x-a)]

notice if you take limit of ln[(x+a)/(x-a)] as x-> infinity, you will get ln1 = 0

xln[(x+a)/(x-a)] = ln[(x+a)/(x-a)]/(1/x)

now the top and bottom will -> 0 so you have an indeterminate and can use l'hopital's.

remember here you're finding the limit of lny

to go from lny to y, use y = e^(lny)

so whatever limit you get for lny, take e to that power

Right. So apply the rule. Don't forget the chain rule...