On rationalising numerator, we get
=limx→∞{√x+1−√x}{√x+1+√x}√x+2{√x+1+√x}
=limx→∞{(√x+1)2−(√x)2}√x+2{√x+1+√x}
=limx→∞{x+1−x}√x+2{√x+1+√x}
=limx→∞√x+2{√x+1+√x} (∞∞ form)
Dividing numerator and denominator by √x, we get
=limx→∞√x+2√x{√x+1+√x}√x
=limx→∞√1+2x{√1+1x+1}
When x→∞, then 1x→0
=√1+2(0){√1+(0)+√1}
=√1{√1+√1}
=12
Therefore,
limx→∞{√x+1−√x}√x+2=12