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In physics, there are often arguments involving infinitesimal quantities. An example from my textbook is the equation $V(x+dx) - V(x) + q(x) dx = 0$. The conclusion is that $q(x) = -\dfrac{dV(x)}{dx}$. This makes sense when treating differentials as fractions, but I was wondering what the formal justification is. This is what I've tried (I replaced $dx$ with $h$ for clarity):

$$V(x+h) - V(x) + q(x)h = 0$$ $$\implies \text{lim}_{h\to0} \ h\cdot \dfrac{V(x+h) - V(x)}{h} + q(x)\cdot h=0$$ $$\implies \dfrac{dV(x)}{dx} \cdot \text{lim}_{h\to0}h = -q(x) \cdot \text{lim}_{h\to0}h$$ and I would get the desired result if I could divide both sides of the equation with $\text{lim}_{h\to0}h$. However, I'm not sure if you can do this, for the same reason $0\cdot 3 = 0 \cdot 5$ does not mean $3=5$.

So is this justified or not, and why or why not? If not, how can I reach my textbook's conclusion in a formally justified way?

Thanks in advance.

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    $\begingroup$ Are you sure the $+V$ wasn't meant to be $-V$? $\endgroup$
    – J.G.
    Nov 14, 2019 at 20:36
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    $\begingroup$ Yes, you're right. Thank you for pointing that out. $\endgroup$
    – Sudera
    Nov 15, 2019 at 11:13

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Assuming that $V(x)$ is differentiable we have

$$V(x+\Delta x)=V(x)+V'(x)\Delta x+o(\Delta x)$$

with $o(\Delta x) \to 0$ and therefore

$$V(x+\Delta x) - V(x) + q(x) \Delta x =V'(x)\Delta x+q(x) \Delta x+o(\Delta x)=0$$

and in the limit this implies

$$V'(x)=-q(x)$$

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  • $\begingroup$ Taking the limit of $\Delta x$ to zero for the equation $V'(x)\Delta x+q(x) \Delta x+o(\Delta x)=0$, the $o(\Delta x)$ would disappear but you still have the $\text{lim}_{\Delta x \to 0} \Delta x$ which make the same problem I asked in my original question: is it justified to divide them out, even if they're approaching zero? $\endgroup$
    – Sudera
    Nov 15, 2019 at 11:21
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    $\begingroup$ @Sudera Yes we can divide by $\Delta x$ and take the limit to obtain the result since $o(\Delta x)/\Delta x\to 0$. $\endgroup$
    – user
    Nov 15, 2019 at 14:51
  • $\begingroup$ I see. Thank you for your answer! $\endgroup$
    – Sudera
    Nov 15, 2019 at 15:05
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    $\begingroup$ You are welcome! Bye $\endgroup$
    – user
    Nov 15, 2019 at 15:09
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Assuming a correction I proposed in a comment:

Use the little-o notation. The original statement is an abbreviation for $V(x+\delta x)-V(x)+\delta q=o(\delta x)$ for small changes $\delta x,\,\delta q$ in $x,\,q$, so $\frac{\delta q}{\delta x}=\frac{V(x)-V(x+\delta x)}{\delta x}=-V^\prime(x)+o(1)$, by the definition of $V^\prime$. In the $\delta x\to0$ limit, $\frac{dq}{dx}=-V^\prime(x)$ by the definition of $\frac{dq}{dx}$.

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