prove $\lim\frac{a_n}{b_n} = \frac{\lim a_n}{\lim b_n}$

I don't understand why the following statement "The trick is to look far enough out into the sequence $(b_n)$ so that the terms are closer to b than they are to 0, which is the main idea of the proof. I also don't understand why is $|b_n| > \frac{|b|}{2}$ ? • Because $|b_n-b|>|b|-|b_n|$ – Zack Ni Aug 15 '16 at 23:09
• why is this inequality true ? – user329017 Aug 16 '16 at 0:53
• If x>y>0, then $|x-y| = x-y = |x|-|y| > |y|-|x|$. If x>0>y, then $|x-y| = x-y > |x|-|y|$ and $x-y > |y|-|x|$ – Zack Ni Aug 16 '16 at 1:03

The idea is that we know that $\left|b_n - b\right|$ can be made smaller than any positive $\varepsilon$ if $b_n \to b$. However, we need to know that the denominator itself does not become too small. Since $b_n \to b$, we can choose some $N_1 \in \mathbb N$ such that $n \geq N_1$ implies that the difference between $b$ and the terms $b_n$ are less than half the distance from $0$ to $b$, and closer to $b$ than $0$. As Arthur points out, the particular bound is chosen for convenience.
• why is $|b_n|$ larger than $|b|/2$? – user329017 Aug 16 '16 at 0:54
• The midpoint between $0$ and $b$ is $b/2$. We have chosen $b_n$ to be closer to $b$ than to $0$. That is, We have chosen the difference $|b - b_n|$ to be smaller than the difference $|b - b/2| = |b/2|$, which is the same as saying that $|b - b_n| < |b|/2$. So, we have chosen $b_n$ to lie to the right (or the left, as the case may be) of $b/2$, hence in absolute value, $|b_n| > |b|/2$. – Alex Ortiz Aug 16 '16 at 1:06
They are using the fact that $b_n$ converges to $b$. By definition, that means that whatever bound you set, sooner or later $b_n$ will be closer to $b$ than that bound. They choose the bound $|b|/2$, for convenience. Thus the definition of convergence tells them that from some point on (i.e. far enough along the sequence), the distance from $b_n$ to $b$ will be less than $|b|/2$. That specifically means that $|b_n|$ must be larger than $|b|/2$.
• why is $|b_n|$ larger than $|b|/2$? – user329017 Aug 16 '16 at 0:54
• Because otherwise it would be further than $|b|/2$ away from $b$. – Arthur Aug 16 '16 at 7:27