Can anyone name the calculus result?

[JonJ678]

Hi. I'm sure there's a named theorem involved here, but I can't remember the name and can't find it through Google. So, it seems worth a shot here.

Given a function f(x) = g(x)/h(x), I'd like to know something about the limiting value of f(x) as x tends to zero.

g(x) tends to zero as x tends to zero. h(x) also tends to zero as x tends to zero. They have no common factor. I'm sure there is a result that states that f(x) will tend to a finite value as x tends to zero.

Can anyone name that theorem?

Many thanks.

 

[Morbius]

L'Hopital's rule.

If g(x) and h(x) tend to 0 as x tends to 0 then f(x) tends to g'(x)/h'(x).

 

[touch]

edit: as above

 

[JonJ678]

I might be able to work with that, but strictly speaking I can't differentiate g or h with respect to x.

The function I'm working with is (efficiency * enthalpy) / speed^2 as speed goes to zero. Enthalpy will tend to a finite limit as speed goes to zero, but efficiency as (work in / useful work) does nasty things as "useful work" is zero at or below low speed.

I'll hold out hope for a result from calculus/analysis which doesn't require the derivative of denominator or numerator. Otherwise I may need to argue that while efficiency is ill defined at low speeds, it probably has a gradient of zero - it's been ill defined since useful work hit zero at higher-than-zero speeds. That's not a solid argument though.

edit: I don't need to know what the limiting value is, only whether it's finite or not. The premise is that efficiency becomes very large at low speeds, while speed becomes very small at low speeds. The authors of this work seem to think that efficiency as something divided by zero will become better behaved if divided by another thing that tends to zero. That doesn't make much sense to me though.

edit2: while the result I was hoping for would still be interesting, it isn't going to save this expression. The numerator tends to infinity, no amount of dividing by x is going to help as x tends to zero. Will have to ask my boss I think

 

[Amleto]

seems a bit of a waste of time since you assert that your numerator is known to be zero, and therefore the value of the limit must also be zero if the denominator is well behaved for values greater than zero.

 

[JonJ678]

Well, I'm fairly sure the enthalpy should be zero at zero speed, as no work is done. There are losses, so it can be constant and non-zero if it needs to be. Isentropic efficiency is badly conditioned, possibly infinite at zero speed. The denominator is definitely zero at zero speed. So I sort of have 0 * ? / 0^2, I'm not convinced I can make a credible argument that it's finite. I don't think I can argue it's infinite either, I don't have the mathematical knowledge to do so.

 

[touch]

the value of the limit must also be zero

consider the simple case of x/(x * x)...

The example you give doesnt have a limit of zero though?
as x approaches zero the result increases.

 

[Duff-Man]

I might be able to work with that, but strictly speaking I can't differentiate g or h with respect to x.

If "g" and "h" are truly non-differentiable in the limit of x -> 0 ; i.e. they are non-smooth functions, then the limit does not exist.

... Unless you just mean that you don't know the functions analytically but that they are still smooth (and therefore differentiable)? In this case, simply asses the gradient of each function wrt x numerically (if you can generate pointwise data in the limit of x-> 0, then a quick and dirty blast in an Excel spreadsheet should give you an approximation to the gradient). If you don't have enough data-points to accurately assess their gradients on approach to zero, then you need more data-samples. If you can't obtain more data samples, then your limit becomes unknown (...or subject to large uncertainties quantified by the degree of uncertainty in your gradient approximations).

I'll hold out hope for a result from calculus/analysis which doesn't require the derivative of denominator or numerator

Unfortunately, math does not work that way... You can reformulate the equation, but the result will always be equivalent to using the original model. In this case you're interested in the behaviour of two functions on approach to a limit - and no matter what way you dress it, you cannot assess limits without examining their derivatives on approach to the limit. If you can't find those derivatives analytically then you must do so numerically.

 

[Amleto]

The example you give doesnt have a limit of zero though?
as x approaches zero the result increases.

I was going to use that as a counter example to the ops wishes, but there is a common factor...


I think there HAS to be a common factor for the limit to be finite, but I'm not sure on that.


Anyway, op says numerator is zero in the neighbourhood of x=0 anyway, so if the denom is well behaved, then no matter about anything else, the limit exists and is 0 as well.

 

[JonJ678]

If "g" and "h" are truly non-differentiable in the limit of x -> 0 ; i.e. they are non-smooth functions, then the limit does not exist.

Unfortunately, math does not work that way... You can reformulate the equation, but the result will always be equivalent to using the original model. In this case you're interested in the behaviour of two functions on approach to a limit - and no matter what way you dress it, you cannot assess limits without examining their derivatives on approach to the limit. If you can't find those derivatives analytically then you must do so numerically.

That's helpful, thank you. The issue is that one of the parameters of concern, isentropic efficiency, is not differentiable everywhere. Specifically at zero speed there is a discontinuity. A related issue is that another parameter, enthalpy, does a poor job of describing the physical reality at low speeds. The machine's behaviour is conveniently continuous, the issue lies with the behaviour of the model. I'll think about refining the model, cheers.

It may be possible to show that both g and h are discontinuous at zero speed, which presumably demonstrates that the limit of g/h is not finite.

 

[Duff-Man]

It may be possible to show that both g and h are discontinuous at zero speed, which presumably demonstrates that the limit of g/h is not finite.

Well it's a little tricky if you want to get technical:

If the discontinuity (in either g or h) occurs at x=0, then there is no single-valued analytical limit (though you may in some cases be able to find two limiting values by examining the limit from the positive and negative side depending on the type of discontinuity). If the discontinuities occur close to, but not strictly at, x=0, then you can still assess by examining the smooth region close to x=0.



... In practice though, when dealing with real physical problems as you are (rather than pure mathematical curiosities), the problem usually comes down to the mathematical model itself breaking-down in the limit - that is, the model no longer adequately represents the physics in the limit.

There is absolutely nothing wrong in stating that the current model is not suitable for (say) low-speed operation. In fact, being precise about the range of applicable scenarios for your model is the most important first step of any scientific modelling. Very few mathematical models are applicable in all scenarios, and so it's to be expected that a different model may be required in certain limits.

If you look around then you may find that there is a different, more suitable model available to estimate efficiency at very low-speed operation. Either way, if you can't measure efficiency accurately for low-speed operation then I wouldn't hold out too much hope for finding an accurate limit in the speed -> zero case.

 

[Amleto]

It may be possible to show that both g and h are discontinuous at zero speed, which presumably demonstrates that the limit of g/h is not finite.

Just to reiterate on some of what duff has said, That quote is totally incorrect.

A discontinuity AT x=0 means nothing with regard to whether a limit exists. The value AT x=0 is totally disregarded when calculating the limit.

A limit can only exist if the converging value is finite, so to talk about a non-finite limit is nonsensical. The limit just does not exist in that case.

 

[JonJ678]

... In practice though, when dealing with real physical problems as you are (rather than pure mathematical curiosities), the problem usually comes down to the mathematical model itself breaking-down in the limit - that is, the model no longer adequately represents the physics in the limit.

...

If you look around then you may find that there is a different, more suitable model available to estimate efficiency at very low-speed operation. Either way, if you can't measure efficiency accurately for low-speed operation then I wouldn't hold out too much hope for finding an accurate limit in the speed -> zero case.

That's gone a long way towards clarifying my thinking on this, thank you. In particular the model needs to be defined so that it breaks down predictably in the limit, such that other physical behaviour can be exposed. In this case, I need to alter the definition of efficiency such that it tends to zero as speed tends to zero- a more physically sensible result than the efficiency tending to +ve infinity as speed approaches zero. Adjusting the model seems a reasonable step to take.

Just to reiterate on some of what duff has said, That quote is totally incorrect.

A discontinuity AT x=0 means nothing with regard to whether a limit exists. The value AT x=0 is totally disregarded when calculating the limit.

A limit can only exist if the converging value is finite, so to talk about a non-finite limit is nonsensical. The limit just does not exist in that case.

Well, perhaps. I don't see a practical distinction between a limit that doesn't exist and a non-finite limit. That's just another sign that my grasp of mathematics is as tenuous as that of most engineering students. I take your point regarding the irrelevance of the function value at x=0 however, thanks for pointing that out explicitly.

Cheers guys, you've been more helpful than I hoped for.

 

[Amleto]

"I don't see a practical distinction between a limit that doesn't exist and a non-finite limit."

It's just nomenclature a non-finite limit is gibberish in maths-talk.