material science - 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!