Let $a_n \rightarrow a$. Show that $\liminf(a_n-b_n)=a-\limsup(b_n)$. Assignment:

Let $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be two sequences of real numbers with $a_n \rightarrow a \in \mathbb{R}$. Show that:
  $$\liminf(a_n-b_n)=a-\limsup(b_n)$$

How to prove this? I have tried to use the identity $\liminf (a_n) = -\limsup(a_n)$ and the triangle inequality, but it didn't work out.
I'd appreciate any help.

Call $B= \limsup_n b_n$.
For all $\varepsilon >0$, eventually you have $$b_n \leq B + \varepsilon \ \ \ , \ \ a - \varepsilon \leq a_n \leq a + \varepsilon$$
So
$$ a- B - \varepsilon\leq a- \varepsilon - b_n \leq a_n - b_n $$
hence $\liminf_n (a_n - b_n) \geq a-B$.
To show that equality holds take a subsequence $b_{n_k} \to B$, so
$$a_{n_k} - b_{n_k} \to a-B$$
so $\liminf_n (a_n - b_n) \leq a-B$.
