What's the solution of a differential equation, when the "limit" is 0 Sorry if the questions sounds horrible to all the mathematicians' ears out there, but my math level is limited, and I just try to get a better intuitive idea of what happens in that case.
I "understand" the concept of differential. It's the change of a function over a certain value (for example the change of speed over time): dV / dt and dt tends to 0. Now my question is, what happens when dt = 0, is it a valid case (can this happens? can it be said), and if yes, is it somehow connected to the Dirac delta function. Finally a friend of mine told me this has somehow a connection with something called the "limit theorem" in differential calculus? I found some info about a "limit theorem" but that doesn't seem to be connected at all to this.
Your help is greatly appreciated, and sorry again if I am not using the right terminology. Please correct me, I am here to learn.
Thank you -
 A: Consider the curve $y = x^2$, and the point $(2,4)$ on that curve. What's the slope of that curve at that point?
Well, if I have two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope of the line connecting them is $\dfrac{y_2-y_1}{x_2-x_1}$. But of course we only have one point here. Let's move along the curve very slightly: let's say that we move from $x = 2$ to $x = 2+\delta$ (where $\delta$ is something really small), so that $y (= x^2)$ is now $(2+\delta)^2$, i.e. $4 + 4\delta + \delta^2$. (Draw yourself a picture of these two points on the curve, and the line through these points.)
The slope between the points $(2,4)$ and $(2+\delta, 4+4\delta+\delta^2)$ is $\dfrac{(4+4\delta+\delta^2) - 4}{(2+\delta) - 2}$, which you can easily simplify to $4+\delta$. So what's the slope at $(2,4)$? Well, it's what happens when $\delta$ tends to 0 (this is formally called a limit) - that is, we'll get $4$. But can we just set $\delta = 0$ right from the start? No - because then we get a 'slope' of $\dfrac{4-4}{2-2}$, which obviously doesn't make any sense.
Now do this again for two points $(x, x^2)$ and $((x+\delta), (x+\delta)^2)$. The slope function you get at the end is called the derivative of $y = x^2$, and often written $dy/dx$, but these $d$ things are just formal symbols, and you shouldn't think of them as numbers. It's just code for what I wrote above.
Does this help?
