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well x=/=0

i would say limit x>0 \[(-\infty,0)\]\[(0,\infty)\]

but im a nub so dont trust me much

well the denominator can never be 0

so x cannot be 0

not much to guess at really

just means if x were zero

the value would be undefined

hence a hole in your graph

got it! ok i got one more...

y=sqrt(2x+3)

well since it is a square root is can never be negative

so it would be \[(0,\infty)\]

the graph of a square root function looks like

|dw:1328895361125:dw|

the books answer is, all X greater than or equal to -3/2

yea thats right

because

y=sqrt(2x+3)

to make the square root 0 x would have to be -3/2

hence its still not a negative

been up all night sorry

getting tired

but since -3/2 make it the sqrt of 0 thats the limit

all other numbers have toi be greater then -3/2

so if u plug in -3/2 "sqrt(2(-3/2)+3" this will equal 0?

yes because the 2's cancel

2x-3/2 = -3

-3+3 = 0

basically

and if you cannot make it undefined

its continuous from -inf to +inf

ok well you have a good one, gotta get some rest now

thanks!