We know that lim f(x)g(x) = lim eg(x)[f(x) -1] if f(x)->1 and g(x) -> ∞ .Actually this is applicable in [1∞ form]
Hence we have,
limx ->∞ [(x + 4)/(x + 3) ]x+1 = limx ->∞ e(x+1)[ (x+4)/(x+3) - 1] since for x->∞ (x+4)/(x+3) ->1 & x+1 -> ∞ . Hence [1∞ form]
= limx ->∞ e(x+1)[ (x+4-x-3)/(x+3)] = limx ->∞ e(x+1)[ 1/(x+3)] = e1 since for x->∞ ,(x+1)/(x+3) -> 1
= e