Approximations in deriving infinitesimal operator commutation relationship I'm having trouble understanding some derivation from Sakurai's QM, chapter 1. To derive $[x,J(dx)]=dx$ he claimed that $dx|x+dx\rangle \simeq dx|x\rangle$ because the approximation is of second order in dx. I don't understand the meaning of this justification...Also, aren't $|x+dx\rangle$ and $|x\rangle$ orthogonal kets? So how come we can even do this kind of approximation?
 A: I'm not sure I see what the problem is, it's explained quite clearly in the section. The equation you have (Equation 1.6.24 in the Revised Edition) basically says:
$$\Big[\mathbf{x}, \mathcal{T}(\text{d}\mathbf{x}^{'})\Big]| \mathbf{x}'\rangle = \text{d}\mathbf{x}^{'} | \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle,$$
and since you know that $$| \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle = \mathcal{T}(\text{d}\mathbf{x}^{'})| \mathbf{x}^{'}\rangle.$$
Now, Sakurai chooses the form of the infinitesimal translation operator to be (in Equation 1.6.20) $$\mathcal{T}(\text{d}\mathbf{x}^{'}) = \mathbb{I} - i \mathbf{K}\cdot \text{d}\mathbf{x}^{'},$$
and so $$\text{d}\mathbf{x}^{'}\mathbb{I}| \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}\mathcal{T}(\text{d}\mathbf{x}^{'})| \mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}\mathbb{I} \left(\mathbb{I} - i \mathbf{K}\cdot \text{d}\mathbf{x}^{'}\right)| \mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}|\mathbf{x}^{'}\rangle + \mathcal{O}(\text{d}\mathbf{x}^{'2})|\mathbf{x}^{'}\rangle,$$
as argued. Now, as to why one can write the Translation operator in that fashion, the chapter in Symmetries will probably help you understand that better. However, it's not too unreasonable. Clearly, as the infinitesimal translation gets smaller and smaller, the translation operator must approach the identity operation (translating something by nothing is equivalent to acting on it by the identity operator).
As to orthogonality, as the other answer points out, the states $|x\rangle$ and $|x+\text{d} x\rangle$ are indeed orthogonal. It can be shown that $$\langle x | x+\text{d} x\rangle = \delta(\text{d}x),$$ which is zero for all non-zero $\text{d}x$.
The claim being made is that the two states are equal only up to the leading order in $\text{d}x$.

EDIT: In case you are interested in how the above relation may be proved, here's a back-of-the-envelope calculation if you are willing to admit that the translation operator is given by $$\mathcal{T}(\text{d}x) = e^{-i \hat{P} dx},$$ where $\hat{P}$ is the momentum operator, and is just $\hbar \mathbf{K}$. (I'll be using units where $\hbar = 1$, and will probably miss a couple of pesky 2$\pi$s.) By definition,
$$\langle x | x  + \text{d}x\rangle = \langle x | \mathcal{T}(\text{d}x) | x\rangle = \langle x | e^{- i P \text{d}x} | x\rangle = \langle x | \sum_{n}{\frac{(- i P \text{d}x)^n}{n!}} | x\rangle.$$
The states of definite position can be decomposed in the momentum basis as $$|x\rangle = \int_{-\infty}^\infty e^{i p x} |p\rangle,$$ and so
\begin{align}
\langle x | \sum_{n}{\frac{(- i P \text{d}x)^n}{n!}} | x\rangle &= \sum_{n}\frac{(-i \text{d}x)^n}{n!}\int \int \text{d}p' \text{d}p\,\, e^{-i (p'-p) x} \langle p'|\hat{P}^n|p\rangle\\
&= \sum_{n}\frac{(-i \text{d}x)^n}{n!}\int \int \text{d}p' \text{d}p \,\, e^{-i (p'-p) x} p^n \delta(p-p')\\
&= \int \text{d}p\,\,\sum_{n}\frac{(-i \text{d}x)^n}{n!} p^n\\
&= \int \text{d}p\,\, e^{-i p \text{d}x}\\ &= \delta (\text{d}x).
\end{align}
A: 
aren't $|x+dx⟩$ and $|x⟩ $ orthogonal kets?

Yes of course they are. But they are actually not equal to each other but it's an approximation of order $dx$.
You can relate it in the following way: Consider a simple Taylor expansion of a function $f(x)$.
$$f(x_0+\Delta x)=f(x_0)+\left |\frac{df}{dx} \right|_{x=x_0}\Delta x+\cdots $$
Now we know that n-tuples can be represented as a vector with n components and in the limit, we can represent a function of ,say $x$, with inifinite number of basis $|x\rangle$. So that the function $f(x)$ can be represented as vector $|f\rangle$ in $|x\rangle$ basis. Refer to Mathematical introduction for Quantum mechanics to any text (or see R.Shankar).
Now writing the taylor series in vector notation, We have
$$|f(x_0+\Delta x)\rangle=|f(x_0)\rangle+\left |\frac{d|f\rangle}{dx} \right|_{x=x_0}|\Delta x\rangle+\cdots$$
For $|x+dx\rangle$ we can do the same
$$|x+dx\rangle=|x\rangle+\left |\frac{d|x\rangle}{dx} \right|_{x}| dx\rangle+\cdots$$
multiplying with $dx$ and taking the term with first order in $dx$ we have:
$$dx|x+dx\rangle\approx dx|x\rangle$$
