Transversal wave speed derivation for small amplitudes 
The above is a derivation for the wave speed equation in my physics textbook. However, I've read online that this equation is only true for waves with small amplitudes. I do not see where this assumption is made in the derivation, so why is the equation only true for small amplitudes?

The above picture shows the vertical restoring force should be 2*T*sin(phi)
 A: The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation. 
The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that $\mathrm{d}y/\mathrm{d} z$ is small, so that the normal to the tangent to the curve stays approximately vertical in the diagram. The best way to understand all this is to work out a more accurate equation; then the vertical component of the force restoring the small length $\mathrm{d}\,s$ of string is 
$$T\,\partial\theta \,\cos\theta = \mathrm{d}s\, T\,\frac{\partial\theta}{\partial s}\,\cos\theta = \mathrm{d}s\, T\,\frac{\partial^2y}{\partial z^2}\,\frac{1}{\left(1+\left(\frac{\partial y}{\partial z}\right)^2\right)^2}$$
(recalling that $\frac{\partial\theta}{\partial s}$ is the curvature of the string and then using the formula for the curvature) and THEN you approximate that $\frac{\partial y}{\partial z}\ll 1$ and, equivalently, that $\mathrm{d}z = \mathrm{d}s$. The small amplitude approximation is then indirect: we're directly assuming small gradients, which imply and are implied by small amplitudes, given that we know the wavelength is limited.
A: Right off the bat in eq (16-23) it's assumed that the restoring force is linear in the displacement.   That's only true for small displacements.
