How do limits fit in with how we derive the differential of a function?

52 Views Asked by At

For example, when I was reading one of my A-level math books, I saw the following:

$\delta A \approx y \ \delta x \ \longrightarrow \ y \approx \frac{\delta A}{\delta x} \ \longrightarrow y=\lim\limits_{\delta x \to 0} (\frac{\delta A}{\delta x} )=\frac{dA}{dx}$

I understand this, but what would happen if we'd taken the limit one step sooner?

$\delta A \approx y \ \delta x \ \longrightarrow \lim\limits_{\delta x \to 0} (\delta A)=\lim\limits_{\delta x \to 0} (y \ \delta x)$

Would this lead to $0=0$ ?

$\\$

My question actually originated when I was reading about circular motion; it was when I saw the following differential:

$ \frac{d}{dt}(\hat{e_\theta)}=-\omega \hat{e_r}$

$\\$

I tried to prove this, and I ended up with the following:

$\Delta\theta \ \hat{e_r} \approx -\Delta \ \hat{e_\theta} $

I was able to show that, as $\Delta\theta $ approached zero, the more accurate this approximation would become(as $\Delta\theta $ approaches zero so does $\Delta \ \hat{e_\theta}$) ; however, I had no idea how to represent this mathematically, I tried writing it as follows:

$\lim\limits_{\Delta\theta \to 0} (\Delta\theta \ \hat{e_r} )=\lim\limits_{\Delta\theta \to 0} (-\Delta \ \hat{e_\theta}) \quad [1]$

I wasn't happy with this representation (since I thought it meant $0=0$), so I changed it into:

$\lim\limits_{\Delta\theta \to 0} (\frac{\Delta\theta}{\Delta t} \ \hat{e_r} )=\lim\limits_{\Delta\theta \to 0} (-\frac{\Delta \hat{e_\theta}}{\Delta t}) \quad [2]$

Then I argued that, as long as $\theta$ is a function of time, I can replace $\Delta \theta$ with $\Delta t$ and write:

$\lim\limits_{\Delta t \to 0} (\frac{\Delta\theta}{\Delta t} \ \hat{e_r} )=\lim\limits_{\Delta t\to 0} (-\frac{\Delta \hat{e_\theta}}{\Delta t})\longrightarrow \frac{d}{dt}(\hat{e_\theta)}=-\omega \hat{e_r} \quad [3]$

Does the transition from [1] to [2] make sense? or does [2] still imply that $0=0$? Is equation [3] equivalent to [2]? Can I argue that, since $\Delta t$ approaches 0 as $\Delta \theta $ approaches 0, equation [2] and [3] are the same?

What is the correct mathematical representation of the above information?

1

There are 1 best solutions below

0
On

Writing

$$\lim_{\delta x\to0}\delta A=\lim_{\delta x\to0}y\,\delta x$$

is correct but useless (indeed, $0=0$ leads you nowhere).

But be careful, this does not allow you to write

$$\lim_{\delta x\to0}\frac{\delta A}{\delta x}=\lim_{\delta x\to0}y$$ because the zero can absorb any constant factor.