Question

help with limit question

Answer 1

what I first did was differentiate the top nominator and bottom denominator and I got \[\frac{ 1 }{ \frac{ 2\sqrt{3y+9} }{ 1 } }\] so this will be the same as \[\frac{ 1 }{ 2\sqrt{3y+9} }\]

Answer 2

next I substituted y=0 into the equation and my final answer turns out to be 1/6 but the answer is actually 1/2 can you tell me what I'm doing wrong?

Answer 3

your derivative is wrong

Answer 4

is it? which part?

Answer 5

you can do this without l'hopital, rationalize the numerator

Answer 6

you forgot the chain rule

Answer 7

\[\frac{d}{dx}\sqrt{3x+9}=\frac{1}{2\sqrt{3x+9}}\times 3\]

Answer 8

oh right k, cause i thought using l'hopital rule you can just differentiate them separately like top by itself and then bottom by itself

Answer 9

you do do it separately
but that means top and bottom separately, no quotient rule
you still have to find the derivative of the top and bottom correctly

Answer 10

okok I think I got it, thanks so much.

Answer 11

yw