How do I find the limit of this equation? With the equation : $$-2d-13.3y+19.3t=0,$$
how do I find the limit of $d$ as $y$ approaches zero when $d, y$  and $t$ are converging to $0$  Some additional information that would help: how do I notate it?
At the suggestion of a commenter, some background:
This is a 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 $d$ and time $t$ 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 $0$?
6 m/microsecond=speed of slower object
19.3 m/microsecond=speed of faster object
total distance=2 m
total time=.333 microseconds
$x$=distance traveled by the 6 m/microsecond object between contact
$y$=time elapsed between contact
$d$=distance between 6 m/microsecond object and endpoint
$t$=time remaining before 6 m/microsecond object reaches endpoint
$d-x$=next closest distance between 6 m/microsecond object and endpoint, becoming the new d
$t-y$=next closest time remaining between 6 m/microsecond object and endpoint, becoming the new t
 A: Here is my understanding, and forgive me, but I'm going to use some different notation:
We have two particles (or point size objects) starting from the same point which I will call $x=0$, and traveling along the same path of length $L = 2$:  $P_1$ with constant speed $r_1=19.3\,\mathrm{m}/\mu \mathrm{s}$ and $P_2$ with constant speed $r_2=6\,\mathrm{m}/\mu \mathrm{s}$.
Particle 2 simply travels along the path until it reaches the end at time $T=\frac{1}{3}\mathrm{s}$.
Particle 1 reflects off of the end point of the path, and off of particle 1, each time that it reaches them.
I believe the questions are:

*

*If we let $y_n$ be the time elapsed between each successive times that the particles were at the same location, what is the limit of $y_n$ as particle 2 approaches the end of the path?

*If we let $x_n$ be the distance traveled by particle 2 between each time that particle 1 makes contact with it, what is an equation for $x_n$?

Answer to 1:
As particle 2 approaches the end of the path, the times between contact with particle 1 become smaller and smaller. The particles would make contact infinitely many times (in the purely mathematical model), and the times would converge to zero.
In mathematical notation: $\lim_{n\to\infty}y_n = 0$
Explanation for 1:
If particle 2 makes contact with particle 1 at some location $x<L$, then particle 1 will instantly reflect and travel back to $x=L$, getting there before particle 2. Since particle 2 is not at $x=L$ yet, particle 1 will instantly reflect and travel toward particle 2, making contact some positive time later. Since particle 1 had to travel for some positive time to reach particle 2, that means that they meet at some location $x<L$. Hence everything repeats, ad infinitum.
The total time is $T=\frac{1}{3}\mathrm{s}$, meaning that the infinite sum of all of the times between contacts (which are all positive) converges. That implies that the sequence of times converges to zero.
In mathematical notation: $\forall n\in\mathbb{N}\, y_n>0$ and $\sum_{n=0}^\infty y_n = \frac{1}{3}$ implies $\lim_{n\to\infty}y_n = 0$.
Answer to question 2:
$$y_n = \frac{2L}{r_1+r_2} \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n-1}$$
$$x_n = \frac{2Lr_2}{r_1+r_2} \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n-1}$$
Explanation for 2:
After $\frac{L}{r_1}$ seconds, particle 1 has reach the end of the path, and particle 2 has traveled $\frac{Lr_2}{r_1}$ meters. Then particle 1 reflects back. At that point they are $L-\frac{Lr_2}{r_1}$ meters apart, and traveling toward each other at speed $r_1+r_2$, so the total time between contact is:
\begin{align}
\frac{L}{r_1} + \frac{ L-\frac{Lr_2}{r_1} }{ r_1+r_2 } &= \frac{L}{r_1}\left( \frac{ r_1+r_2 }{ r_1+r_2 } \right) + \frac{L}{r_1}\left( \frac{ r_1-r_2 }{ r_1+r_2 } \right)\\
& = \frac{L}{r_1}\left( \frac{ 2r_1 }{ r_1+r_2 } \right)\\
& =  \frac{ 2L }{ r_1+r_2 } 
\end{align}
After that much time, particle 2 is at $\frac{ 2Lr_2 }{ r_1+r_2 }$, so the distance from particle 2 to the end of the path is:
\begin{align}
L - \frac{ 2Lr_2 }{ r_1+r_2 } = L\left( \frac{ r_1-r_2 }{ r_1+r_2 } \right)
\end{align}
At this point, everything repeats, but the system has contracted by a factor of $\left( \frac{ r_1-r_2 }{ r_1+r_2 } \right)$, so the next time and distance traveled get multiplied by this factor. This repeats ad infinitum, yielding the geometric series for $y_n$ and $x_n$.
The total time $T_n$ after $n$ contacts, and the total distance traveled $X_n$, can be found by summing the geometric series:
\begin{align}
T_n &= \sum_{i=1}^{n}y_i = \frac{2L}{r_1+r_2} \left(\frac{ 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n} }{ 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right) }\right)\\
&=\frac{L}{r_2}\left( 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n} \right)
\end{align}
\begin{align}
X_n &= \sum_{i=1}^{n}x_i = \frac{2Lr_2}{r_1+r_2} \left(\frac{ 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n} }{ 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right) }\right)\\
&= L\left( 1 - \left(\frac{r_1-r_2}{r_1+r_2}\right)^{n} \right)
\end{align}
And we see that as $n\to\infty$, $T_n\to \frac{L}{r_2}$ and $X_n\to L$, as expected.
ADDED
I am not sure how you get $\frac{2d-x}{t-y}=r_1$
If you define $d_n$ to be the distance from the slower particle to the end at the moment of the $n^\mathrm{th}$ contact, and $t_n$ to be the time remaining at that moment until the slower particle reaches the end, then the faster particle travels $d_n + d_{n+1} = 2d_n - x_{n+1}$ distance in a time equal to $y_{n+1}$.
Using the $X_n$ and $T_n$ that I defined above, we have that $d_n = L - X_n$ and $t_n = \frac{d_n}{r_2}$.
If we let $q = \frac{r_1-r_2}{r_1+r_2}$, we showed above that $X_n = L(1-q^n)$, so $d_n = Lq^n$
Using the equation for $y_n$ from above also, this gives us the equation:
$$\frac{d_n + d_{n+1}}{y_n} = \frac{Lq^n + Lq^{n+1}}{ \frac{2L}{r_1+r_2} q^n} = \frac{Lq^n\left(\frac{2r_1}{r_1+r_2}\right)}{ \frac{2Lq^n}{r_1+r_2} } = r_1$$
Which is exactly what it must equal.
In summary, in this purely mathematical model, the faster particle reflects infinitely many times, and as the slower particle approaches the end, both the time between contact and the distance traveled between contacts approaches zero. This can be seen directly in the formulas derived for those quantities.
