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Teachers and Students => Universities => Topic started by: SGVaibhav on January 07, 2010, 04:37:52 pm

Title: Limits question
Post by: SGVaibhav 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

Title: Re: Limits question
Post by: astarmathsandphysics 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./

Title: Re: Limits question
Post by: SGVaibhav 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))

Title: Re: Limits question
Post by: SGVaibhav 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)

Title: Re: Limits question
Post by: astarmathsandphysics on January 07, 2010, 07:36:01 pm

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

Title: Re: Limits question
Post by: SGVaibhav on January 07, 2010, 07:42:05 pm

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

Title: Re: Limits question
Post by: astarmathsandphysics on January 07, 2010, 08:46:30 pm

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

Title: Re: Limits question
Post by: astarmathsandphysics on January 07, 2010, 11:24:17 pm

Here is an explanation of L'Hospital's rule

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

Title: Re: Limits question
Post by: SGVaibhav 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)