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

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:

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



$-3/2<x+1<-1/2 \implies$ ?

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

  • $\begingroup$ I don't know what you mean about wanting two things to be positive. $\endgroup$
    – dfeuer
    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


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


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


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

  • $\begingroup$ @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. $\endgroup$
    – Jared
    Jul 17 '13 at 19:38
  • $\begingroup$ But, in this exercise, can I replace delta by 2, 3, 5, 1/4, 1/9 and so on? or only 1/2? $\endgroup$
    – Tomi
    Jul 17 '13 at 19:44
  • $\begingroup$ 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 $\frac{1}{|x+1|}<\frac{1}{1-\delta}$. $\endgroup$
    – user84413
    Jul 18 '13 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.