Let $(A_i,\Phi_{ij})$ be an inductive system of $C^*$-algebras with inductive limit $A$. Let $B$ be another $C^\ast$-algebra.
I want to show that for the minimal tensor product we have $\mathrm{lim}(A_i\otimes B)=A\otimes B$.
Clearly we have an isomorphism $\mathrm{lim}(A_i\odot B)\cong A\odot B$ on the algebraic tensor product. I want to show that this extends to an isomorphism of $C^\ast$-algebras with respect to the norm of the minimal tensor product on $A\otimes B$ and the induced norm on the limit $\mathrm{lim}(A_i\otimes B)$. In other words: the induced norm on the inductive limit is the norm of the minimal tensor product.
Can somebody help? What does the induced norm look like?
This is probably not true in general and the reason is that the minimal tensor product functor $-\otimes B$ is not necessarily exact. In contrast to that, the maximal tensor product functor $-\otimes_\max B$ is exact, and this allows one to conclude continuity with respect to inductive limits; see this post.
In the case that $B$ is an exact $C^*$-algebra, i.e. $-\otimes B$ preserves short exact sequences, then what you claim is true. Let's see how this works:
Let $A_1\xrightarrow{\phi_1}A_2\xrightarrow{\phi_2}A_3\to\dots$ be an inductive system with inductive limit $(A,\{\mu_n\}_{n\ge1})$. For $m\ge n$, put $\varphi_{m,n}:=\varphi_m\circ\varphi_{m-1}\circ\dots\circ\varphi_n:A_n\to A_m$. Tensoring with $B$, we have the inductive system $$A_1\otimes B\xrightarrow{\phi_1\otimes\mathrm{id}_B}A_2\otimes B\xrightarrow{\phi_2\otimes\mathrm{id}_B}\dots$$ and note that the pair $(A\otimes B,\{\mu_n\otimes\mathrm{id}_B\}_{n\ge1})$ satisfies $(\mu_n\otimes\mathrm{id}_B)=(\varphi_n\circ\mu_{n+1})\otimes\mathrm{id}_B=(\varphi_n\otimes\mathrm{id}_B)\circ(\mu_{n+1}\otimes\mathrm{id}_B)$ for all $n\in\mathbb{N}$. So, all we need to do is verify the following two conditions:
If it is not clear why this suffices, I recommend looking at chapter 6 of Rordam's blue book. The first condition is easily verified by some density arguments. For the second condition, the inclusion of the right-hand-side inside the left-hand-side is easy (why: take $x\in A_n\otimes B$ with $\lim_m\|\varphi_{m,n}\otimes\mathrm{id}_B(x)\|=0$. Then $\|\mu_n\otimes\mathrm{id}_B(x)\|=\|(\varphi_{m,n}\otimes\mathrm{id}_B\circ(\mu_m\otimes\mathrm{id}_B)(x)\|\le\|\varphi_{m,n}\otimes\mathrm{id}_B(x)\|$ for all $m>n$, so letting $m\to\infty$ yields that $\|\mu_n\otimes\mathrm{id}_B(x)\|=0$). For the other inclusion, we use exactness of $B$: let $n\in\mathbb{N}$ and consider the short exact sequence $0\to\ker(\mu_n)\xrightarrow{\iota}A_n\xrightarrow{\pi}\mu_n(A_n)\to0$ (where $\iota$ is the inclusion map and $\pi$ is the corestriction of $\mu_n$, i.e. $\pi:A_n\to\mu_n(A_n)$ defined as $\pi(a)=\mu_n(a)$). By tensoring with $B$, we obtain a short exact sequence $0\to\ker(\mu_n)\otimes B\xrightarrow{\iota\otimes\mathrm{id}_B}A_n\otimes B\xrightarrow{\pi\otimes\mathrm{id}_B}\mu_n(A_n)\otimes B\to0$. By exactness, we have that $\ker(\pi\otimes\mathrm{id}_B)=\text{Im}(\iota\otimes\mathrm{id}_B)$.
Consider the inclusion map $j:\mu_n(A_n)\to A$ (which is injective). Since $\otimes$ preserves injectivity, we have that $j\otimes\mathrm{id}_B:\mu_n(A_n)\otimes B\to A\otimes B$ is injective. Now $\mu_n\otimes\mathrm{id}_B=(j\circ\pi)\otimes\mathrm{id}_B=(j\otimes\mathrm{id}_B)\circ(\pi\otimes\mathrm{id}_B)$, and since $j\otimes\mathrm{id}_B$ is injective, we have that $\ker(\mu_n\otimes\mathrm{id}_B)=\ker(\pi\otimes\mathrm{id}_B)=\text{Im}(\iota\otimes\mathrm{id}_B)$.
It suffices to show that $\iota\otimes\mathrm{id}_B(y)$ belongs to the RHS of condition 2 for all $y\in\ker(\mu_n)\otimes B$ -- and it further suffices to show this for elementary tensors, since these generate the tensor product as a $C^*$-algebra. So let $a\in\ker(\mu_n)$ and $b\in B$. Since $(A,\{\mu_n\})$ is by assumption the inductive limit of $(A_j,\phi_j)$, we have that $\ker(\mu_n)=\{a\in A_n: \lim_{m\to\infty}\|\varphi_{m,n}(a)\|=0\}$. Now $$\|(\varphi_{m,n}\otimes\mathrm{id}_B)(a\otimes b)\|_{A_m\otimes B}=\|\varphi_{m,n}(a)\otimes b\|=\|\varphi_{m,n}(a)\|\cdot\|b\|\xrightarrow[m\to\infty]{}0$$
and we are done.