Show that if $\lim\limits_{j \to \infty} b_j = \beta$ and $\beta < 0$ then $\exists$ $N > \mathbb{N}$ such that j > N $\Rightarrow$ $b_j < 0$.

Hi I'm taking Analysis I.

I'm trying to solve the following:

Show that if $\lim\limits_{j \to \infty} b_j = \beta$ and $\beta < 0$ then $\exists$ $N > \mathbb{N}$ such that j > N $\Rightarrow$ $b_j < 0$.

I intuitively know that this is true but I don't know how to start the proof. Start of with the delta-epsilon definition of limit, but then where to go?

• Hint: Take $\epsilon = \frac{|\beta|}{2}$. – Anu Feb 12 '18 at 4:25
• @Anu, The number was used earlier in my answer below :). – Benicio Feb 12 '18 at 4:26

But there is some $N$ such that $b_{j} - \beta < |\beta|/2$ for all $j > N$ by assumption (try to see this from the $\varepsilon-N$ definition of convergence.). So $b_{j} < \beta + |\beta|/2 = \beta/2 < 0$ for all $j > N$.
$\forall \epsilon > 0$, exist $n_{0} \in \mathbb{N}$ such that $n > n_{0} \Longrightarrow |b_{n} - \beta| < \epsilon \Longrightarrow b_{n} \in (\beta - \epsilon, \beta + \epsilon)$. Take $\epsilon = \frac{-\beta}{2}$, so exist $n_{0} \in \mathbb{N}$ such that $n > n_{0} \Longrightarrow b_{n} \in (\frac{3\beta}{2}, \frac{\beta}{2})$. The result follows.