Showing a proposition of sequence

Question

How would I show the following

If limit

$j\rightarrow \infty$ for the sequence $b_j=B$ and B<0 then there exist an

number N in natural number such that when j>N then $b_j<0$

Would I start off doing

$|b_j-(-B)|<\epsilon$

Answer 1

Since $B < 0$, then $\varepsilon := -B/2 > 0$. So since $\lim_{j\to \infty} b_j = B$, there exists a natural number $N$ such that $|b_j - B| < \varepsilon$ for all $j > N$. Show that $|b_j - B| < \varepsilon$ implies $b_j < B + \varepsilon = B/2$.