# Limit at Infinity of Rational Function

I'm following the Algorithm Design book by Eva Trados. In the 'Basics of Algorithm Analysis' chapter, I've come across a proof that I'm having difficulty understanding and can't find any proper explanation for it.

$$\lim_{n \rightarrow \infty } \frac{f(n)}{g(n)} = c > 0$$ Given two functions f and g, the proof uses the fact that if a positive limit exists for the limit above, then there is some value $n_{0}$ beyond which the ratio is always between $\frac{1}{2}c$ and 2c.

What I don't understand is, how do we know that the ratio is between $\frac{1}{2}c$ and 2c ?

• That is true for any sequence $(a_n)$ with limit $c\gt 0$. May 20, 2014 at 0:32
• I am trying to understand the reasoning behind it. Can you direct me to some reading or book that explains this ? May 20, 2014 at 0:33

Let $(a_n)$ be any sequence with limit $c\gt 0$. Then by the definition of limit, for any $\epsilon \gt 0$, there is an $N$ such that if $n\gt N$ then $|a_n-c|\lt \epsilon$.
Pick $\epsilon=\frac{c}{2}$. Then there is an $N$ such that if $n\gt N$ we have $|a_n-c|\lt \frac{c}{2}$, or equivalently $$c-\frac{c}{2}\lt a_n \lt c+\frac{c}{2}.$$ From this we can conclude that if $n\gt N$ then $\frac{c}{2}\lt a_n\lt 2c$.
• Shouldn't c + $\frac{c}{2}$ be $\frac{3}{2}c$? May 20, 2014 at 0:39
• @WasifHyder Then you'd get $\frac c 2 <a_n<\frac3 2 c$. But $\frac 3 2 c<2c$, so what you want follows. May 20, 2014 at 0:40
• Yes, but if $a_n\lt \frac{3}{2}c$ then $a_n\lt 2c$. May 20, 2014 at 0:41
• You are welcome. We gave a rather formal answer, but the point is that after a while $a_n$ is very close to $c$. We could have chosen $\epsilon=\frac{c}{10}$, and concluded that there is an $N$ such that if $n\gt N$ then $9c/10\lt a_n\lt 11c/10$. May 20, 2014 at 0:44