# Prove a limit using the epsilon-delta definition

$$\lim_{x\to-2}\frac{4x-1}{x+1} = 9$$

Given $$\epsilon>0$$, $$(\exists \delta(\epsilon)>0) \left( |x+2|<\delta \implies \left|{\frac{4x-1}{x+1} - 9}\right| < \epsilon \right)$$

So, if $$|x+2|<\delta$$, then: $$\left|{\frac{4x-1}{x+1} - 9}\right| = \left|{\frac{4x-1-9(x+1)}{x+1}}\right| = \left|{\frac{-5x-10}{x+1}}\right| = \left|{\frac{-1 (5x+10)}{x+1}}\right| = {\frac{|-1||5x+10|}{|x+1|}} = {\frac{5|x+2|}{|x+1|}}$$

I've found the expression $$5|x+2|$$ in the numerator:

$$|x+2| < \delta = 1/2$$

$$-1/2

$$-5/2

$$-3/2 ?

A colleague told me to leave both members of this inequality as positives. Why do we need both sides to be positives?

• I don't know what you mean about wanting two things to be positive. Jul 17 '13 at 18:55

You're right to want to bound values of $x$ away from $-1$, and to do so by $\frac{1}{2}$ is fine. The inequality you obtain by requiring $x$ to be within $\frac{1}{2}$ of $2$ is

$$-\frac{5}{2}<x<-\frac{3}{2}$$

which you have written. Then, as you've written, we have

$$-\frac{3}{2}<x+1<-\frac{1}{2}$$

Notice that the above inequality implies that $|x+1|>\frac{1}{2}.$ It follows that

$$\frac{5|x+2|}{|x+1|}<10|x+2|$$

So let $\delta(\epsilon)<\min\{\frac{1}{2},\frac{\epsilon}{10}\}$.

• @TomiSebastiánJuárez: There is no rule about what signs are allowed in inequalities. Something like $-1<x<1$ makes perfect sense, it is just stating that $x$ lies in between the numbers $-1$ and $1$. However, it is true that $1<x<-1$ has no solutions, because a number cannot be greater than $1$ and less than $-1$ at the same time. Jul 17 '13 at 19:38
• But, in this exercise, can I replace delta by 2, 3, 5, 1/4, 1/9 and so on? or only 1/2?
– Tomi
Jul 17 '13 at 19:44
• You could replace 1/2 by any number a with $0<a<1$, since then $0<|x+2|<\delta$ implies that $|x+1|>1-\delta$ and therefore $\frac{1}{|x+1|}<\frac{1}{1-\delta}$. Jul 18 '13 at 0:23