# Tricky epsilon-delta proof

$$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$

What I've got so far is that:

$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\ \forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$$

How do I go about finding a value for delta from here? Thanks.

• @Mason apologies, that was a typo on my part. Oct 8 '19 at 7:23

Take the case where $$x\in [-2,0]$$ (i.e. $$\delta < 1$$). Then we have the following inequality

$$\left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1|$$

by maximizing the numerator and minimizing the denominator. So set $$\delta = \min(\frac{3}{5}\epsilon,1)$$ and the proof step for the limit follows in both cases.

If $$\epsilon > \frac{5}{3}$$:

$$|x+1|<1\implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \frac{5}{3} < \epsilon$$

If $$\epsilon \leq \frac{5}{3}$$:

$$|x+1|<\frac{3}{5}\epsilon \implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \epsilon$$

• Everything seems great except I don't understand where you get the interval of [-2,0]. Could you elaborate on that, please? Thank you. Oct 8 '19 at 15:53
• Edit: I've figured out how you got [-2,0] but why do we need to get the greatest possible value out of the fraction if we want the minimum for our delta anyway? Oct 8 '19 at 16:18
• @mathgeek101 I wrote out the proof step because that's what makes it clear why we maximize the fraction. Oct 8 '19 at 18:38

Note that you have $$0<|x+1|<\delta$$, and you have $$(x+1)$$ in your numerator in the second line. So we make this more explicit: $$\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|=|x+1|\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\epsilon$$ This is what we want to hold. Now, in order to get a handle on this, we have $$|x+1|\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\delta\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|$$ and $$\delta$$ we can control. So we want to pick a $$\delta$$ small enough that the right-hand side here is less than $$\epsilon$$, since that will automatically make the left-hand side smaller than $$\epsilon$$. Thus we have to see how large that second factor on the right-hand side could possibly become. I claim that it's always smaller than $$1$$. Which in turn gives us: $$\delta\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\delta$$ So as long as we pick $$\delta= \epsilon$$, putting together all these inequalities gives us $$\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$$, which is what we want, and we have proven our limit.

Proof of claim: We can either do calculus to find max and min, or we do some algebra and split into cases. I'll go with the algebra option. Note that the numerator $$3x^2-3x+3$$ is always positive, so we may remove the absolute value signs from it. This gives us $$\left|\frac{2x-1}{3x^2-3x+3}\right| = \frac{|2x-1|}{3x^2-3x+3}<1\\ |2x-1|<3x^2-3x+3$$ For $$x\leq \frac12$$, this turns into $$1-2x<3x^2-3x+3$$, which is easily verified by the quadratic formula. For $$x\geq\frac12$$, it turns into $$2x-1<3x^2-3x+3$$, which is also easily verified with the quadratic formula. So we get that $$\left|\frac{2x-1}{3x^2-3x+3}\right|<1$$, and we are done.