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Limits question
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Topic: Limits question (Read 2009 times)
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
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astarmathsandphysics
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Re: Limits question
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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./
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SGVaibhav
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Re: Limits question
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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))
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SGVaibhav
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Re: Limits question
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Reply #3 on:
January 07, 2010, 07:33:03 pm »
O.o
SIR I GOT THE ANSWER BEFORE U
(saying it with pride)
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astarmathsandphysics
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Re: Limits question
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Reply #4 on:
January 07, 2010, 07:36:01 pm »
I was right but should be 2/(1/(2*-3))
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SGVaibhav
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Re: Limits question
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Reply #5 on:
January 07, 2010, 07:42:05 pm »
fractions sound difficult over here, better show it in paint or something..
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astarmathsandphysics
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Re: Limits question
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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
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astarmathsandphysics
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Re: Limits question
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Reply #7 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
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SGVaibhav
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Re: Limits question
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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)
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