How to show the limit of finite differences is equal to the differential?

Question

Consider the relationship between the density and volume of a substance:

$$V \rho = m$$ or $$V = m\rho^{-1} \tag{1}$$

If we take the differential of both sides while holding the mass constant, we get:

$$dV = -\frac{m}{\rho^2}d\rho \tag{2}$$

Now the question comes in with the change in volume as related to the change in temperature. This relationship is given by:

$$dV = \alpha VdT \tag{3}$$

where alpha is a material constant. If we simply substitute (2) and (1) into (3) we end up with:

$$d\rho = -\alpha \rho dT\tag{4} $$

But say we write the finite difference version of (3) as:

$$\Delta V = \alpha V_1 \Delta T \tag{5}$$

Now from (1) we get:

$$\Delta V = V_2 - V_1 = \frac{m}{\rho_2}-\frac{m}{\rho_1} = \alpha \frac{m}{\rho_1} \Delta T \tag{6}$$

This leads to:

$$\Delta \rho = -\alpha \rho_2\Delta T \tag{7}$$

Alternatively, we could start with:

$$\rho_2 = \frac{m}{V_1 + \Delta V} = \frac{\rho_1V_1}{V_1 + \Delta V} \tag{8}$$

Now using (5) in (8) we get:

$$\rho_2 = \frac{\rho_1}{1+\alpha \Delta T} \rightarrow \rho_2-\rho_1 = \Delta \rho = -\frac{\rho_1 \alpha \Delta T}{1+\alpha \Delta T} \tag{9}$$

So we have two expressions for the finite difference in density, (7) and (9). My question is how to show that these both go to (4) in the limit of smaller and smaller finite differences?

I think in the case of (7) this is fairly easy because as the finite difference gets smaller and smaller,

$$\Delta \rho \rightarrow d\rho , \Delta T\rightarrow dT, \rho_2 \rightarrow \rho \tag{10}$$

But I can't think of a how the last relation in (10) is justified rigorously. Is there a limit rule that says this? I am not sure how to proceed in the case of (9) because it seems if I say the term in the denominator goes to 0, so that the denominator goes to 1, then how can I avoid the numerator going to zero as well?

Answer 1

for a function T, $$\Delta T = T(a+x)-T(a) = dT(x) + O(x)$$ where O(x) is defined as ||O(x)||/||x|| going to zero as x goes to zero, and dT is a linear function.

If T is one variable, dT(x) is simply dT times x. Thus $$\ \frac{\Delta T}{x} = \frac{T(a+x)-T(a)}x = \frac{dT(x)}x + \frac{O(x)}x = dT + \frac{O(x)}x$$ dT is a number, T'(a).

Then Δp = −ρ1αΔT/1+αΔT, looking at T and p as the identity function T(t1)=t1

(dp + O)(1+α(dT + O)) = −ρ1α(dT + O)

dp + αdpdT + O + OO + OαdT

a property of O is that O times a linear function is O and a bounded linear function times a bounded linear function is O

Thus, dp = −ρ1αdT + O

note that the differential of a linear function is the function itself: T(a+x)-T(a) = dT(x) + O(x) = T(x)

your question can also be answered by the fact that (lim f) x (lim g) = lim gxf

so lim −ρ1αΔT/1+αΔT = (lim −ρ1αΔT)/(lim 1+αΔT).

you can "choose" to make the denominator 1, since it is a different limit, and you can "choose" to maintain the numerator as (lim −ρ1αΔT), which goes to −ρ1αdt