Translation Operator 
Let $|\psi\rangle \to |\psi'\rangle = \hat{T}(\delta x)|\psi\rangle$ for infinitesimal $\delta x.$ Show that $\langle x \rangle' = \langle x \rangle + \delta x$ and $\langle p_x \rangle' = \langle p_x\rangle.$

I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x?$
Shouldn't it equal $\langle x \rangle?$ Since, $\langle x\rangle' = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|x\hat{T}(\delta x)|\psi\rangle$ and using $\hat{T}(\delta x) = e^{-i\hat{p}_x\delta x/\hbar}$ then $\langle \psi |\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle = \langle x\rangle.$
 A: 
I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x$?

Because you acted with the translation operator on the state. This is by definition what we want the translation operator to do. If it doesn't do this then we are in trouble.

Shouldn't it equal $\langle x \rangle?$ 

Nope.

Since, $\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle =\langle \psi'|x\hat{T}(\delta x)|\psi\rangle$ 
  and using
  $\hat{T}(\delta x) = e^{-i\hat{p}_x\delta x/\hbar}$ then 
  $\langle \psi|\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle = \langle{}x\rangle.$

Nope. 
$$
T^\dagger x T\neq T^\dagger T x
$$
so
$$
\langle \psi'|x\hat{T}(\delta x)|\psi\rangle
\neq
\langle \psi|\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle
$$
You should have:
$$
\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle =\langle \psi'|x\hat{T}(\delta x)|\psi\rangle= \langle \psi|\hat{T}^\dagger(\delta x)x\hat{T}(\delta x)|\psi\rangle
$$
Then you have to commute $T$ and $x$, and then use the fact that $T^\dagger T=1$.
HINT[!!!]:
$$
[x,p]=i\hbar
$$
