Limits question

[SGVaibhav]

Tried this weird question for 1 hours, still i could not get the answer..
Attached the question

[astarmathsandphysics]

L'hopitals rule again.
f=sin2x and f(0)=0
g=3-sqrt(x+9) and g(0)=0
so f/g=f'/g'=(2cos2x)/(1/(2sqrt(x+9))=2/6=1/3
I did a page on Lhospitals rule and will put it up tonight./

[SGVaibhav]

correct answer is -12
i think there is a mistake in differentiation
u wrote f/g = f'/g'
it should be    (f'g  - g'f)/(g'^2 (square))

[SGVaibhav]

O.o

SIR I GOT THE ANSWER BEFORE U (saying it with pride)

[astarmathsandphysics]

I was right but should be 2/(1/(2*-3))

[SGVaibhav]

 
fractions sound difficult over here, better show it in paint or something..

[astarmathsandphysics]

write it as 2/(1/-6)=(2/1)/((1/-6) then use the dividing fractions rule

[astarmathsandphysics]

Here is an explanation of L'Hospital's rule

http://www.astarmathsandphysics.com/university_maths_notes/calculus/university_maths_notes_calculus_lhospitals_rule.html

[SGVaibhav]

yeah so it is (2/1)/((1/-6)           , so i can change the division to multiplication, so it will become            2  x   -6
By the way, in  ur page u have written  L hospitals rule
its L hopitals rule


and in the 4th rule, u have written   f" (a)    /      g"  (a)