As an example, suppose I want to prove that the limit of $f(x)=x^3+5x^2-2$ when $x\to 1$ is $4$. A solution given by Mark Viola is:
Note that for $0<|x-1|<1$
$$\begin{align} \left|x^3+5x^2-6\right|&=|x^2+6x+6||x-1|\\\\
&<22|x-1|\\\\ &<\epsilon \end{align}$$whenever $|x-1|<\delta=\min\left(1,\frac{\epsilon}{22}\right)$
It is not quite clear to me why the minimum is employed. I understand that setting $\delta \leq 1$ is typically done. Is this because there are no inherent restrictions on $\delta$? Is the minimum $\delta=\min\left(1,\frac{\epsilon}{22}\right)$ set so that in case we are left with a $\delta>1$, then we can fall back onto $\frac{\epsilon}{22}$?
If so, this doesn't seem to make sense as $\frac{\epsilon}{22}$ was constructed under the $\delta \leq 1$ condition. Instead, should I read the $\min\left(1,\frac{\epsilon}{22}\right)$ condition as saying "both conditions must hold?"
Best Answer
$0 <\delta < \min (1,\frac {\epsilon} {22})$ is equivalent to $0<\delta <1$ and $\delta <\frac {\epsilon} {22}$.
It is easy to make $|x^{2}+6x+6||x-1| <\epsilon$ if we have a bound for $|x^{2}+6x+6|$: If $M$ is a bound for this then $|x-1| <\epsilon /M$ would give $|x^{2}+6x+6||x-1| <\epsilon$. To get such a bound we take $\delta <1$ and observe that $|x| \leq |x-1|+1<\delta+1 <2$ if $|x-1| <\delta$. This helps us to get a bound $M$.