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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 [1form]

 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 

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