Made mistake *** NEW Basic calculus limit question FML save me will rate quickly

im really sorry i made a mistake
y approaches 1 for this question not zero like the previous one that i asked
sorry everyone please answer this one

lim y-->1 (y-(4sqrt(y)) +3) / (y^2+1)

how in the hell do i do this

Putting y = 1 we get
=> (1-(4sqrt(1)) +3) / (1^2+1)
= Now sqrt(1) = 1 or sqrt(1) = -1
=> (1-(4sqrt(1)) +3) / (1^2+1) or (1+ (4sqrt(1)) +3) / (1^2+1)
= (1-4+3)/2 or (1+4+3)/2
= 0 or 4

lim{y→1} [y - 4√(y) + 3] / (y² + 1) = [1 - 4 + 3]/(1 + 1) = 0/2 = 0

looks to me like at y = 1, top is 0, bottom is 2, so limit is 0. don't even need L'Hopital.