Problem:
Let $|\psi' \rangle = \hat T(\delta x) | \psi \rangle$ for infinitesimal $\delta x$, show that $\langle x \rangle'= \delta x +\langle x \rangle$
Attempt at Solution:
Using $\hat T(\delta x) = 1-\frac{i}{\hbar} \hat P_x \delta x$ and $[\hat x, \hat P_x] = i \hbar$, I calculate $[\hat x, \hat T(\delta x)]=\delta x$
Therefore,
$$\hat x \hat T(\delta x)=\delta x + \hat T(\delta x) \hat x$$
Plug this into $\langle \psi | \hat T^{\dagger}(\delta x) \ \hat x \ \hat T(\delta x) | \psi \rangle$
$$\langle \psi | \hat T^{\dagger}(\delta x) (\delta x + \hat T(\delta x) \hat x) | \psi \rangle$$
$$\langle \psi | \hat T^{\dagger}(\delta x) \delta x | \psi \rangle + \langle \psi | \hat T^{\dagger}(\delta x) \hat T(\delta x) \hat x | \psi \rangle$$
Since $\hat T$ is unitary, this almost simplifies to the result:
$$\langle \psi | \hat T^{\dagger}(\delta x) \delta x | \psi \rangle + \langle \hat x \rangle$$
Question:
What am I doing wrong here? I've seen on wikipedia that $[\hat x, \hat T(\delta x)]=\hat T(\delta x) \delta x$, which would solve the problem since
$$\langle \psi | \hat T^{\dagger}(\delta x) \ \delta x \ \hat T(\delta x) | \psi \rangle$$
$$\delta x \langle \psi | \hat T^{\dagger}(\delta x) \hat T(\delta x)| \psi \rangle$$
$$\delta x $$
but I'm really not seeing how to get there with the way Townsend has defined things.