Basic calculus limit question FML save me will rate quickly

[From: ] [author: ] [Date: 11-10-07] [Hit: ]

.please help me-you should use lhopitals rule. Which involves you taking the derivative of the top and bottoms of the quotient independently as if they were seperate functions. the first time should yield (1-2*y^(-1/2))/2*y and then once more and your down to y^(-3/2)/2 and that should be a fairly simple limit.= 3-If you plug in zero for y, you just get 3/1 = 3.......

lim y-->0 (y-(4sqrt(y)) +3) / (y^2+1)

how in the hell do i do this
ive tried everything...conjugate factoring...i keep getting 0

please help me

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you should use l'hopital's rule. Which involves you taking the derivative of the top and bottoms of the quotient independently as if they were seperate functions. the first time should yield (1-2*y^(-1/2))/2*y and then once more and your down to y^(-3/2)/2 and that should be a fairly simple limit.

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Simply put y = 0
=> (0-(4sqrt(0)) +3) / (0^2+1)
= 3/1
= 3

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If you plug in zero for y, you just get 3/1 = 3.

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hmmm

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Maby 0 is the answer. Only in 9th grade so I would not know.

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