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Limits question

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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 :D :P :P :D :D :D (saying it with pride)

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

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