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Topic: Another Sequence (Read 2833 times) |
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rmsgrey
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Re: Another Sequence
« Reply #25 on: Apr 24th, 2007, 6:34am » |
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on Apr 24th, 2007, 12:42am, towr wrote:
Isn't it 1 everytime sin(x)=0 ?
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For that matter, isn't it bounded between e and 1/e?
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towr
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Re: Another Sequence
« Reply #26 on: Apr 24th, 2007, 8:05am » |
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on Apr 24th, 2007, 6:34am, rmsgrey wrote:| For that matter, isn't it bounded between e and 1/e? |
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No, because (x + cos x sin x) can grow unbounded. I think you missed the x term there.
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Eigenray
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Re: Another Sequence
« Reply #27 on: Apr 24th, 2007, 1:20pm » |
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on Apr 23rd, 2007, 1:41pm, Aryabhatta wrote:| I think Eigenray is pointing to the fact that it seems odd to use LHospital's rule to just see that the limit is f'(0). The given limit is the definition of f'(0). |
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Right. It's not wrong, exactly, just backwards. (I don't hate l'Hopital as much as some people.)
A classic example is: by l'Hopital,
limx->0 sin(x)/x = sin'(0)/1 = cos(0) = 1,
even though the standard proof that sin'(x) = cos(x) uses the fact that sin(x)/x -> 1.
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| « Last Edit: Apr 24th, 2007, 1:23pm by Eigenray » |
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Eigenray
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Re: Another Sequence
« Reply #28 on: Apr 24th, 2007, 1:36pm » |
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on Apr 23rd, 2007, 5:05pm, Icarus wrote:
on Apr 24th, 2007, 12:42am, towr wrote:
Isn't it 1 everytime sin(x)=0 ?
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on Apr 24th, 2007, 6:34am, rmsgrey wrote:
For that matter, isn't it bounded between e and 1/e? |
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on Apr 24th, 2007, 8:05am, towr wrote:
No, because (x + cos x sin x) can grow unbounded. I think you missed the x term there.
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I assume you're all talking about the ratio
f(x)/g(x) = e-sin(x) (for x  0).
It oscillates between 1/e and e, hence it diverges (because it doesn't converge). I'm confused by the confusion.
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towr
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Re: Another Sequence
« Reply #29 on: Apr 24th, 2007, 2:59pm » |
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Oops, I seem to have switched from considering the ratio to considering just g(x)
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Icarus
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Re: Another Sequence
« Reply #30 on: Apr 24th, 2007, 5:51pm » |
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I got that example from the Wikipedia article on L'Hopital's rule. Though it seems towr is now past his confusion, let me spell it out for others:
f(x) = x + cos x sin x = x + (1/2)sin 2x
g(x) = esin x(x + cos x sin x) = esin xf(x)
f(x)/g(x) = e-sin x, which oscillates back and forth between 1/e and e as x   , so f/g diverges.
f'(x) = 1- sin2 x + cos2 x = 2cos2 x
g'(x) = cos x esin x(x + cos x sin x) + esin x(2cos2 x) = cos x esin x(x + cos x sin x + 2cos x).
f'/g' = (2cos x)/esin x(x + cos x sin x + 2cos x).
Since the top is bounded while the bottom grows without bound (because of the x term), limx f'/g' = 0.
This shows that L'Hopital's Rule can fail when the denominator is allowed to be 0 in every neighborhood of the point. (To apply the same example at finite points, you can compose it with a transformation such as x = 1/|t|.)
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| « Last Edit: Apr 24th, 2007, 5:51pm by Icarus » |
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And e is just as cursed.
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When their digits are reversed? " - Anonymous
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