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?

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 $|f(x)-f(2)| < \varepsilon$.
