# Can you divide (identical) limits on both sides of an equation?

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?

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

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

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

• 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? Nov 15, 2019 at 11:21
• @Sudera Yes we can divide by $\Delta x$ and take the limit to obtain the result since $o(\Delta x)/\Delta x\to 0$.
– user
Nov 15, 2019 at 14:51
• I see. Thank you for your answer! Nov 15, 2019 at 15:05
• You are welcome! Bye
– user
Nov 15, 2019 at 15:09

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 $$\frac{dq}{dx}$$.