Author Topic: Limits question  (Read 2009 times)

Offline SGVaibhav

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Limits question
« on: January 07, 2010, 04:37:52 pm »
Tried this weird question for 1 hours, still i could not get the answer..
Attached the question

Offline astarmathsandphysics

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Re: Limits question
« Reply #1 on: January 07, 2010, 05:07:07 pm »
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./

Offline SGVaibhav

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Re: Limits question
« Reply #2 on: January 07, 2010, 05:28:59 pm »
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))

Offline SGVaibhav

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Re: Limits question
« Reply #3 on: January 07, 2010, 07:33:03 pm »
O.o

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


Offline astarmathsandphysics

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Re: Limits question
« Reply #4 on: January 07, 2010, 07:36:01 pm »
I was right but should be 2/(1/(2*-3))

Offline SGVaibhav

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Re: Limits question
« Reply #5 on: January 07, 2010, 07:42:05 pm »
 ???
fractions sound difficult over here, better show it in paint or something..

Offline astarmathsandphysics

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Re: Limits question
« Reply #6 on: January 07, 2010, 08:46:30 pm »
write it as 2/(1/-6)=(2/1)/((1/-6) then use the dividing fractions rule

Offline astarmathsandphysics

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Re: Limits question
« Reply #7 on: January 07, 2010, 11:24:17 pm »

Offline SGVaibhav

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Re: Limits question
« Reply #8 on: January 08, 2010, 07:55:39 am »
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)