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Find the values of a and b so that the following is true.

Mathematics
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I am thinking about L'Hospital's rule but I am not sure how I would apply it...
well, you can separate them, then solve them
Yeah.... But I have two variables then.

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Other answers:

are you sure it tends to infinity?
Yep.
0.
oops.
why two variables?
if you differentiate the variable you get nothing from a, but b will still be present, you can get b right?
\[\lim_{x \rightarrow 0} (\frac{ \sin(2x) }{ x^3 }+a+\frac{ b }{ x^2 })=0\]
Yeah. I think you are right.
ahhh, so it tends to zero?
Yep :P . I typed wrong.
no problem, you can separate them then do the limits, you can spot it by inspection, very easily
Hmm Let me try...
have you done it?
I can't solve the limit of: \[\lim_{x \rightarrow 0}\frac{ \sin(2x) }{ x^3 }\]
you can!! differentiate 3 times until the denominator is 1,
not 1, differentiate until the denominator is a constant
sin(2)=2sin(x)cos(x)
I got -4/3 .
for?
The limit.
sin(2x)/x^3 as x tends to infinity?
how ?
To 0 :P .
lol!!! im losing it today...
but still, i would have thought it was zero
Waitt.
The Limit is infinity..... :( .
yep,
We can't solve anything then.
we could, i mean if one of the variables was infinity
But We can't do anything with infinity.... It's not a number.
wait are you sure it should tend to 0?
Well I would assume it should tend to a finite value.
i suspect the limit should tend to infinity then for it so satisfy the limit
not necessarily, it can be infinity, let assume it was infinity, then we can do the question with ease. have you done analysis by any chance?
No.....
Here is the question.
1 Attachment
okay, no problem, assume it is infinity because i think you were right the first time round, if it was infinity just plug in infinity to the differential you had computed
If we did that then a would be infinity and b would not exist.
b would, it would be x^2,
one moment,
but since b^2 tends to infinity the whole thing would be 0.
x^2 sorry not b^2.
which would be 0, at when b/x^2 as x ->0
Exactly. We can't work with that.
we could, if b = 0,
0/anynumber = 0
But it's not. B/x^2 is zero, not b.
We don't know what b is.
a could be -infinity, then this would satisfy the limit
But infinity- infinity is not 0.
yes it is
No it's not. Infinity is not a number. It's a concept.
that is true, i moment, i am trying to multi task, i jst remembered the proof that it isnt,
http://en.wikibooks.org/wiki/Calculus/Infinite_Limits/Infinity_is_not_a_number
okay, you need to apply l'hpitals rule over and over again, i would recon, but first you need to put all the equation under one common factor so \[=\frac{\sin(2x) + ax^{3} + bx}{x^{3}}\]
http://answers.yahoo.com/question/index?qid=20110406112720AARpKn1 here is a better and easier way of doing it the sint is taylor expansion i think, the rest is pretty self explanatory
Lol. THanks :P .

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