(x+a)n=xn+naxn-1+... (binomial expansion)
(x+a)1+1/x=x1+1/x+(1+1/x)ax1/x+...
When x→∞, 1/x→0, so (x+a)1+1/x→x+a (as you might expect, because the exponent→1).
x1+1/(x+a)→x as x→∞, so (x+a)1+1/x-x1+1/(x+a)→x+a-x=a.
l'Hôpital's Rule requires quotients of two functions f and g with their derivatives. We don't have the quotient of two functions.
Also, (x+a)1+1/x=(x+a)(x+a)1/x=x+a, x1+1/(x+a)=x.x1/(x+a)=x as x→∞. Hence the difference is a.