Fractional change of density I'm asked to prove that the fractional change of density of a fluid ($\frac{\Delta\rho}{\rho_0}$) is so that $$\frac{\Delta\rho}{\rho_0}=-\beta\,\Delta{T},$$ where $\beta$ is the volumetric coefficient of expansion and given that $\Delta{V}=\beta\,V_0\,\Delta{T}$. However, my attempt at a solution, which starts from $\Delta{\rho}$: 
$$
\Delta{\rho}=m\,\left[\frac{1}{V_0+V_0\,\beta\,\Delta{T}}-\frac{1}{V_0}\right]=\frac{-\beta\,\Delta{T}}{1+\beta\,\Delta{T}}\,\rho_0
\implies\frac{\Delta\rho}{\rho_0}=\frac{-\beta\,\Delta{T}}{1+\beta\,\Delta{T}}.
$$
ends up in something different. Did I do something wrong? or is it the problem that's poorly written?
 A: Your analysis is not wrong. Note that the general definition of the coefficient of expansion is $\beta \equiv \frac{1}{V} \frac{dV}{dT}$; so when you write it as $\beta \simeq \frac{1}{V} \frac{\Delta V}{\Delta T}$, you're implicitly assuming that the temperature variation $\Delta T$, and hence $\Delta V/V$ is small${}^*$.
With that in mind, using $\Delta V = \beta V \Delta T$, your final equation can be rewritten as:
$$\frac{\Delta \rho}{\rho_0} = - \frac{\beta \Delta T}{1+\Delta{V}/V}$$
As mentioned previously, the change in volume due to expansion is much less than the initial volume of the object so $\Delta V/V << 1$, meaning that you can neglect $\Delta V/V$ in the denomenator in comparison with $1$; resulting in:
$$\frac{\Delta \rho}{\rho}\simeq-\beta \Delta T$$

* To be more precise, from the definition of $\beta$ we have:
 $$\frac {dV}{V}=\beta dT$$
so:
 $$\int_{V_1}^{V_2}\frac {dV}{V}=\int_{T_1}^{T_2}\beta dT$$
Now provided that the temperature variation $\Delta T = T_2 - T_1$ is sufficiently small so that we can neglect the variations of $\beta$ in the $[T_1,T_2]$ interval, $\beta$ (and $V$) can be approximately pulled out of the integral, resulting in:
$$\frac{\Delta V}{V} \simeq \beta \Delta T$$
which is the formula you were using.
A: Note that you can also rearrange the definition of $\beta$ in terms of the density $\rho=\frac m V$ to get that approximation:
$$\beta \equiv \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p} = \frac{1}{m / \rho}\left(\frac{\partial (\frac {m}{\rho})}{\partial T}\right)_{p} = \frac {\rho}{m}\left(-\frac{m}{\rho^2}\right)\left(\frac{\partial \rho}{\partial T}\right)_{p}=-\frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}\right)_{p}$$
So
$$\beta \simeq-\frac{1}{\rho} \frac{\Delta \rho}{\Delta T} \implies \frac{\Delta\rho}{\rho}=-\beta\,\Delta{T}$$
