Let $b \in \mathbb{R}$. How do I show that $\frac{b}{n}_{n \in \mathbb{N}}$ → 0?

Question

How do I solve this problem? So far, I've done: Take $\epsilon > 0$. $|\frac{b}{n}-0| = |\frac{b}{n}|$. For $\epsilon > 0$, there exists $N \in \mathbb{N}$ s.t $\frac{1}{N} < \epsilon$ (by Archimedian property). For all $n > N$, $|\frac{b}{n}-0|=|\frac{b}{n}| \leq \frac{b}{n} \leq \frac{1}{N} < \epsilon \implies$ $\frac{b}{n} \rightarrow 0$. But clearly there is a problem in this last step. Thanks!

Answer 1

If $b=0$, then this is obvious. Otherwise, for the Archimedian property, instead of taking $N\in\mathbb{N}$ s.t. $\frac{1}{N}<\epsilon$, try taking $\frac{1}{N}<\frac{\epsilon}{|b|}$. This gives $\frac{|b|}{N}<\epsilon$, and you've got the rest.

Answer 2

The biggest problem with the last step is if $b>1$, then $\frac{b}{n}\not\leq \frac1n$, which means we can't conclude that $\frac bn\leq \frac1N$. Your definition of $N$ has to take $b$ into account in some way, not just $\epsilon$.