In mathematics, a limit is the value that a function approaches as the input approaches some value. You are starting with an equation. Would it help to solve for x as a function of y and z, y as a function of x and z, z as a function of x and y? I can't see that you have given what the limit or limits would be...seems like some things are missing. Some classification of the type of math you are doing might help, just to look up references or example problems.
This is a philosophy forum so you're doing sort of a Hail Mary Pass here. There might be one or two good math guys or gals in the bunch, but it's not me.

I hope this thread isn't too banal for the forum, but I've got a math problem related to some physics concepts that requires calculus and I'm not sure how to solve it. How do you find the limit of the equation -2x-13.3y+19.3z=0, with x, y and z greater than zero? I'm not even sure where to begin with a three variable equation, so any help you guys can provide will be great! The math savvy can probably teach me and additional posters something about calculus. — Enrique

If you can state the math problem verbatim, I'll help you with it if I can. As it stands you've not stated a calculus problem. Try writing using Mathjax if you have very long expressions, it also helps display limits clearly, eg:

$\displaystyle{\lim\limits_{x\to \infty} \frac{1}{x} = 0}$.

Seems like a unusual question for one who started threads like "Qualia and Quantum Mechanics, The Sequel" and "Do Physics Equations Disprove the Speed of Light as a Constant?"

It doesn't make sense as posed. What you describe is a plane in 3-space. To see it , graph at 3D Grapher

I think this is a valid description of the problem. Two entities leave the same point at the same time within a fixed path length but at different, constant rates, and each time they meet the slower rate entity continues at the same pace while the faster rate entity resumes its travel forward and backward between contact and the endpoint at the same pace. If total distance of the path length is 2 and total time elapsed is 0.333, at what distance and time from the endpoint will y be zero?

Equation of the distance covered and time elapsed between contacts at the faster rate is (2d−x)/(t−y)=19.3. Each d−x value becomes the new d value in the next iteration, and each t−y value becomes the new t value. Equation of the slower rate is x/y=6. What is the limit of d as y approaches 0?

The math here keeps shifting. I thought the plane in 3 dim was the best answer now everything has changed with iterations thrown in. Convince me you're not jerking our chains.

Trying not to jerk your chain! I'm uncertain what the range of geometrical representations possible for the problem is, probably multiple ways of modeling it spatially.

Ok, you're new to me, but looks like you've been around awhile.

Ok, you're new to me but looks like you've been around awhile.

You still have not clearly explained the problem. Tell us exactly what x, y, t, d, and z represent. It sounds a bit like the old bee vs train problem where the bee keeps going back and forth at a constant rate between the moving train and the wall at the end of the track. That has a simple resolution.

-2d-13.3y+19.3t=0 — Enrique

Try substituting f(m)=2^d!-Piy*183*(m!)^2.5 where m= 3.3540606060...

This will make d and y equal to -2 and plus 44, respectively,

Which reduces your original equation to stupidity,

Which makes Enrique fall out both sides,

And you get 3=1, which is the basis of the Christian numbering system.

probably multiple ways of modeling it spatially. — Enrique

Send one of those models over to me, I need some jerking around here.

Enrique is a genius. He's invented the sport of math guy chain jerking. It might catch on with high school students. Ultimately a good thing that keeps the math guys on their toes.

Great, wise guys lol

You still have not clearly explained the problem. Tell us exactly what x, y, t, d, and z represent. It sounds a bit like the old bee vs train problem where the bee keeps going back and forth at a constant rate between the moving train and the wall at the end of the track. That has a simple resolution. — jgill

That's exactly what it is, lets put it in terms of bees and trains. The bee moves at 19.3 m/microsecond and the train at 6 m/microsecond (really fast bee and train).

x=distance traveled by the train between contact
y=time elapsed between contact
d=distance between train and wall
t=time remaining before the train reaches the wall
d-x=next closest distance between train and wall, becoming the new d
t-y=next closest time remaining between train and wall, becoming the new t

Again, the equations as I've figured them so far are:

(2d-x)/(t-y)=19.3 m/microsecond

x/y=6 m/microsecond

-2d-13.3y+19.3t=0

Total (initial) distance d is 2 m and total (initial) time t is 0.333 microseconds. The distance x and time y between contacts gets smaller, and I'm trying to find the distance from the wall at which x and y = 0, which I've estimated as slightly larger than x=.0052 m.

If you know a solution to this variation of the bee/train problem, describe it to me!

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The bee/train problem is slightly different from my recollection, but basically the same ideas:Bee vs Trains
The process doesn't end before the trains collide, or, in my version, before the train hits the wall.

You seem to be knowledgeable about the resources available in this area of study: do you know of a cheap or free application for making good quality scientific graphs and inserting them into a document?

I write all my graphics with Liberty BASIC language, then insert the images in Word documents, then convert to pdf. I'm not familiar with other programs or services. Others who post on this forum might know of something.

The matplotlib library in Python and the ggplot2 library in R, both open source, are useful for custom plotting. For extreme customisation (and corresponding workload per plot) you can't beat the tikz library in LaTeX though. The latter's more of a high powered doodling tool, the former two are data visualisation libraries.