Limits of Poisson's ratio in isotropic solid For an isotropic solid, Poisson's ratio can be expressed in terms of stiffness constants as:
$$\sigma = \frac{c_{11} - 2c_{44}}{2c_{11} - 2c_{44}}$$
Alternatively we may express Poisson's ratio in terms of the Lamé constants $\lambda$, $\mu$ where $\lambda = c_{12}$ and $\mu = c_{44}$.  For an isotropic solid, we have $c_{12} = c_{11} - 2c_{44}$.  When we solve this algebraically we get:
$$\sigma = \frac{\lambda}{2(\lambda + \mu)}$$ 
I am supposed to show that $\sigma$ must lie between $-1$ and $+\frac{1}{2}$.  I figure that by setting $\mu=0$, it follows from the last expression of $\sigma$ that we get the value $\frac{1}{2}$.  However, I can't see how to show that the lower bound must be $-1$.  I've tried fidgeting around with the expressions and letting them go towards $0$ or $\infty$, but I still don't get the $-1$ value.  If anyone can help me out here, I would greatly appreciate it!
 A: This is going to be a bit of a long answer, so please bear with me while I develop the explanation fully. You can use energy arguments to prove the bounds on the Lamé parameters (and therefore, the Poisson ratio). Here is the free energy function for a linear elastic solid:
$$
\psi = (1/2)\mathbf{\epsilon}:\mathcal{\mathbf{C}}:\mathbf{\epsilon}
$$
where $\mathbf{\epsilon}$ is the infinitesimal strain tensor, and $\mathcal{\mathbf{C}}$ is the 4th order elasticity tensor. The components of the elasticity tensor are
$$
C_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \mu (\delta_{ik}\delta_{jl}
+\delta_{il}\delta_{jk})
$$
where $\delta_{ij}$ is the Kronecker delta. The energy can be written in terms of these components (with implied summation over repeated indices).
$$
\psi = C_{ijkl}\epsilon_{ij}\epsilon_{kl}=2\mu |\epsilon|^2+\lambda(\mathrm{tr}\:\epsilon)^2
$$
The strain can be broken into its spherical part and deviatoric part $\epsilon'$
$$
\epsilon'=\epsilon -\frac{1}{3}\mathrm{tr}(\epsilon)\mathbf{1}
$$
Now re-write the energy in terms of the deviatoric and spherical components of strain. This is the result after some algebra
$$
\psi = 2\mu|\epsilon'|^2 + (\lambda+\frac{2}{3}\mu)(\mathrm{tr}\:\epsilon)^2
$$
The key idea that we use to prove the bounds on physical constants is that the energy $\psi$, which represents the stored energy due to mechanical deformation, must be positive. That is, $\psi \geq 0$.
Now, imagine a deformation which is PURELY deviatoric. This implies that $\mathrm{tr}\:\epsilon = 0$. Therefore, the energy is now
$$
\psi = 2\mu |\epsilon'|^2 > 0
$$
Since the magnitude of a tensor is always a positive number, this places the restriction that $\mu > 0$. 
Next, imagine that you have a purely dilitational deformation. This implies that $\epsilon' = \mathbf{0}$. Therefore, the energy for this deformation is
$$
\psi= (\lambda+\frac{2}{3}\mu)(\mathrm{tr}\:\epsilon)^2 > 0
$$
Therefore, we have the restriction that $\lambda+\frac{2}{3}\mu > 0$.
The parameters $\mu, \lambda$ are not used often in engineering calculations, so they can be combined in various ways to form other elastic constants. In fact, $\mu$ is the shear modulus, which is often denoted as G. The quantity $\lambda+\frac{2}{3}\mu$ is known as the bulk modulus (K), or the stiffness of a material in response to a volume change.
Now, to actually answer your original question, we need to formulate Poisson's Ratio in terms of the quantities G and K. Rather than go through the pages of algebra to do so, I will direct you to a table at the bottom of this page
http://en.wikipedia.org/wiki/Young%27s_modulus
and give you the answer.
$$
K(G, \nu) = \frac{2G(1+\nu)}{3(1-2\nu)} > 0
$$
and
$$
G(K,\nu) = \frac{3K(1-2\nu)}{2(1+\nu)} > 0
$$
Clearly, to satisfy both of these constraints (and consequently, to ensure that the quantities G,K are bounded), the following restrictions must be placed on $\nu$
$$
-1 < \nu < \frac{1}{2}
$$
