<s>[INST] 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?
 [/INST] 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 $|x+3|<6.$
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