Continuity for $f(x)=x^2-x$ 
Let $f:\mathbb{R} \to \mathbb{R},$ $f(x)=x^2-x.$ Prove using the definition that $f$ is continuous at $-1$.

So $|f(x)-f(-1)| = |x^2-x-2| = |x-2||x+1|$.
I cannot seem to get past this point. How can i find $\delta$ for this? 
 A: The "trick" is to bound $|x+1|$ by choosing an initial value of $\delta>0$.  Let's arbitrarily choose $\delta=1/2$.
Then, $|x+1|<1/2$ implies that $-3/2<x<-1/2$.  Then, we see that $|x-2|<7/2$.  So, we see that given any $\varepsilon>0$, 
$$|x^2-x-2|<\varepsilon$$
whenever $|x+1|<2\varepsilon/7$, provided $\delta<1/2$.  So, choose $\delta =\min\left(\frac12,\frac27 \varepsilon\right)$ and we are done!
A: $\delta$ controls the size of $|x+1|$, so you have to account for the additional term $|x-2|$. However, as $x$ gets closer to $-1$, $|x-2|$ gets closer to $3$. For example, if $|x+1|<1$, then $|x-2|= |(x+1)+(-3)|\leq |x+1|+3<4$. 
With that in mind, given $\epsilon>0$, let $\delta = \min(\epsilon/4,1)$. Then $|x+1|<\delta$ implies $|x+1|<\epsilon/4$ and $|x+1|<1$. The latter implies $|x-2|<4$, so coupling with the former implies
$$|x^2-x-2| = |x+1||x-2| < (\epsilon/4)\cdot 4 = \epsilon$$
A: For any $\epsilon>0$, let $\delta=\min\{\epsilon/4, 1\}$. Then if $|x+1|=|x-(-1)|<\delta$, 
$$
|x-2||x+1|<4\delta\leq \epsilon
$$
