Here is a quotation of a book: ($\otimes$ denotes the minimal tensor product)
Lemma 3.9.2. Let $A$ be a C*-algebra. If $E\subset A$ is an operator system and $J\triangleleft B$ is an ideal, then there is an isometric inclusion $$\frac{E\otimes B}{E \otimes J} \subset \frac{A\otimes B} {A\otimes J}$$.
Proof. We must show that if $x\in E\otimes B$, then its norm down in $\frac{A\otimes B}{A\otimes J}$ is equal to $$\inf\{||x+y||: y\in E\otimes J\}.$$
This is easily seen since the norm in $\frac{A\otimes B}{A\otimes J}$ is equal to $$\lim||x(1\otimes(1-e_{i}))||,$$ where $\{e_{i}\}$ is an approximate unit for $J$.
My question is why "the norm in $\frac{A\otimes B}{A\otimes J}$ is equal to $\lim||x(1\otimes(1-e_{i}))||$" above?
By definition, for $x\in A\otimes B$, $$ \|x+A\otimes J\|=\inf\{\|x-y\|:\ y\in A\otimes J\}. $$ Now, for $y\in A\otimes J$, $$ x-y=\lim_j (x-y(1\otimes e_j))=\lim_j\,x(1\otimes (1-e_j))+(x-y)(1\otimes e_j). $$ The elements inside the limit are of the form $x(1\otimes (1-e_j))+z$, $z\in A\otimes J$. Thus $$ \|x+A\otimes J\|=\inf\{\lim_j\|x(1\otimes (1-e_j))+(x-y)(1\otimes e_j)\|:\ y\in A\otimes J\}\\ =\lim_j\|x(1\otimes(1-e_j))\| $$ (note that we can take $y=x(1\otimes e_k)$ for $k$ arbitrarily large).
The inspiration for this is easier for me to see in the von Neumann algebra case. If $M$ is a vN algebra and $J\subset M$ a ultraweakly closed ideal, then $J=pM$ for some central projection $p\in M$. Then $$ \|x+J\|=\text{dist}(x,J)=\inf\{\|x-py\|:\ y\in M\}=\inf\{\|x(1-p)+(x-y)p\|:\ y\in M\}\\ =\inf\{\max\{\|x(1-p)\|,\|(x-y)p\|\}:\ y\in M\,\}=\|x(1-p)\| $$ (note that the infimum is taken over number all greater than or equal $\|x(1-p)\|$, and this number is achieved when $y=x$).