Proof: $f(x)=x^2+x-4$ is continuous at $x=2$ Proof: Let $\epsilon>0$ and $\delta =\min\{1,\epsilon/6\}$.Then if $|x-2|<\delta$, $|x-2|<1$, so $|x+3|<6$.
Thus, $$|f(x)-f(c)|= |x^2+x-4-(2^2+2-4)| = |x^2+x-6|=|x-2||x+3|<6\delta \leq \epsilon$$
Doing this problem as practice. Thoughts? comments?
 A: If $x \in \Bbb{R}$, then $|f(x) - f(2)| = |x^{2}+x-6| = |x-2||x+3|$; we have $|x-2| < 1$ only if $|x| - 2 \leq |x-2| < 1$, only if $|x| + 3 < 6$, only if $|x+3| \leq |x|+3 < 6$, and only if $|x-2||x+3| < 6|x-2|$; given any $\varepsilon > 0$, we have $6|x-2| < \varepsilon$ if $|x-2| < \varepsilon/6$. Hence we have proved: for every $\varepsilon > 0$, we have $|x-2| < \min \{ 1, \varepsilon/6 \}$ only if $|f(x)-f(2)| < \varepsilon$.
A: It seems correct, however this technique relies on you being able to factor the polynomial. That means you could not use your recipe for polynomial of higher degrees. 
A more generic approach is to use the estimate of $|x-2|<\delta$ to get an estimate of $|(x-2)^n|<\delta^n\le\delta$ and then use the triangle inequality:
$$|f(x)-f(2)| = |(x-2+2)^2+x-2-4| = |(x-2)^2 + 4(x-2) + 4 + (x-2) - 4| = |(x-2)^2 + 5(x-2)| \le |(x-2)^2| + 5|(x-2) < 6\delta$$
this technique is basically what's then used to show that sum and products of continuous functions are again continuous (by which result you could prove this trivially).
A: Late answer but let us see if $\delta= \min \lbrace 1,\frac{\epsilon}{6} \rbrace $ works or not.  First case is if $\delta=1.$ If $|x-2|<1,$  then $$|f(x)-f(2)|=|x^2+x-6|=|x-2||x+3|<(1)\left(6\right)<\frac{\epsilon}{6}(6)=\epsilon$$ since $|x-2|<1$ implies both $|x-2| < \frac{\epsilon}{6}$ and $|x+3|<6.$ 
Now let $\delta = \frac{\epsilon}{6}.$ Proceeding like before, we have $$|f(x)-f(2)|=|x^2+x-6|=|x-2||x+3|<\frac{\epsilon}{6}(6)=\epsilon$$ since $|x-2|<\frac{\epsilon}{6}$ implies $|x-2|<1$ and $|x+3|<6.$
