I have come across something which is most likely wrong. If there is a better place to put this, please tell me. Also, please let me know why it is wrong.
Taking the linear slope of the parabola function $f(x) = x^2$ with $\frac{f(x)-f(x-1)}{x-(x-1)}$ (which is the linear slope function $\frac{y_2-y_1}{x_2-x_1}$ with the parabola function fit into it), the value will constantly get larger. $\frac{1}{\frac{f(x)-f(x-1)}{x-(x-1)}}$ ($\frac{1}{\frac{x^2-(x-1)^2}{x-(x-1)}}$ without using the function) will get constantly smaller. If we put $∞$ in for this expression though, the result will be $\frac{1}{\frac{0}{0}}$. $$\frac{1}{\frac{∞^2-(∞-1)^2}{∞-(∞-1)}}$$ $$\frac{1}{\frac{∞-(∞)^2}{∞-(∞)}}$$ $$\frac{1}{\frac{∞-(∞)^2}{∞-(∞)}}$$ $$\frac{1}{\frac{∞-∞}{∞-∞}}$$ $$\frac{1}{\frac{0}{0}}$$
If we use logic (which is probably not a good idea for most things) based in infinity, with something getting smaller and smaller with each larger value of $x$, and since any normal value being able to be represented by some decimal number, and because the slope cannot be a straight line (slope $0$), the slope must be $.000...1$ (infinite zeros). This number is the smallest possible number (so it cannot get smaller with a larger value of $x$, which should be a value that infinity gives), every other number will have a number directly above them that gives a smaller number, and finally, $.000...1$ is the number directly above $0$, all of the previously defined conditions are true.
This expression can be simplified to $\frac{1}{2x-1}$ with: $$\frac{1}{\frac{x^2-(x-1)^2}{x-(x-1)}}$$ $$\frac{x-(x-1)}{x^2-(x-1)^2}$$ $$\frac{x-(x-1)}{2x-1}$$ $$\frac{1}{2x-1}$$
Putting any number in both expressions will give the same result, although when you put $∞$ in for it you will get $1/∞$, which evaluates to $∞$. Obviously, this is because the simplification does not include $∞-∞$.
Despite the above though, evaluating the first expression in two different ways gives $.000...1=\frac{1}{\frac{0}{0}}$, and possibly (but not likely) $.000...1=\frac{1}{\frac{0}{0}}=\frac{1}{∞}$. Why is this?