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

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.

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  • $\begingroup$ @Mason apologies, that was a typo on my part. $\endgroup$ Oct 8 '19 at 7:23
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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 $\epsilon \leq \frac{5}{3}$:

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

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  • $\begingroup$ Everything seems great except I don't understand where you get the interval of [-2,0]. Could you elaborate on that, please? Thank you. $\endgroup$ Oct 8 '19 at 15:53
  • $\begingroup$ 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? $\endgroup$ Oct 8 '19 at 16:18
  • $\begingroup$ @mathgeek101 I wrote out the proof step because that's what makes it clear why we maximize the fraction. $\endgroup$ Oct 8 '19 at 18:38
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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.

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