Dido525
Find the values of a and b so that the following is true.
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Dido525
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I am thinking about L'Hospital's rule but I am not sure how I would apply it...
Rezz5
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well, you can separate them,
then solve them
Dido525
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Yeah.... But I have two variables then.
Rezz5
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are you sure it tends to infinity?
Dido525
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Yep.
Dido525
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0.
Dido525
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oops.
Rezz5
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why two variables?
Rezz5
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if you differentiate the variable you get nothing from a, but b will still be present, you can get b right?
Dido525
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\[\lim_{x \rightarrow 0} (\frac{ \sin(2x) }{ x^3 }+a+\frac{ b }{ x^2 })=0\]
Dido525
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Yeah. I think you are right.
Rezz5
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ahhh, so it tends to zero?
Dido525
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Yep :P . I typed wrong.
Rezz5
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no problem, you can separate them then do the limits,
you can spot it by inspection, very easily
Dido525
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Hmm Let me try...
Rezz5
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have you done it?
Dido525
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I can't solve the limit of:
\[\lim_{x \rightarrow 0}\frac{ \sin(2x) }{ x^3 }\]
Rezz5
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you can!!
differentiate 3 times until the denominator is 1,
Rezz5
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not 1, differentiate until the denominator is a constant
Rezz5
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sin(2)=2sin(x)cos(x)
Dido525
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I got -4/3 .
Rezz5
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for?
Dido525
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The limit.
Rezz5
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sin(2x)/x^3 as x tends to infinity?
Rezz5
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how ?
Dido525
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To 0 :P .
Rezz5
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lol!!! im losing it today...
Rezz5
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but still, i would have thought it was zero
Dido525
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Waitt.
Dido525
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The Limit is infinity..... :( .
Rezz5
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yep,
Dido525
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We can't solve anything then.
Rezz5
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we could, i mean if one of the variables was infinity
Dido525
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But We can't do anything with infinity.... It's not a number.
Rezz5
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wait are you sure it should tend to 0?
Dido525
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Well I would assume it should tend to a finite value.
Rezz5
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i suspect the limit should tend to infinity then for it so satisfy the limit
Rezz5
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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?
Dido525
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No.....
Dido525
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Here is the question.
Rezz5
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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
Dido525
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If we did that then a would be infinity and b would not exist.
Rezz5
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b would, it would be x^2,
Rezz5
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one moment,
Dido525
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but since b^2 tends to infinity the whole thing would be 0.
Dido525
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x^2 sorry not b^2.
Rezz5
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which would be 0, at when b/x^2 as x ->0
Dido525
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Exactly. We can't work with that.
Rezz5
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we could, if b = 0,
Rezz5
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0/anynumber = 0
Dido525
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But it's not. B/x^2 is zero, not b.
Dido525
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We don't know what b is.
Rezz5
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a could be -infinity,
then this would satisfy the limit
Dido525
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But infinity- infinity is not 0.
Rezz5
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yes it is
Dido525
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No it's not. Infinity is not a number. It's a concept.
Rezz5
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that is true, i moment, i am trying to multi task,
i jst remembered the proof that it isnt,
Rezz5
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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
\[=\frac{\sin(2x) + ax^{3} + bx}{x^{3}}\]
Dido525
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Lol. THanks :P .