Why do we need an upper bound for $|x+5|$ when $x$ is close to $3$? Show that $\lim_{x\to 3}$ $x^2 + 2x + 6 = 21$ using the $\epsilon-\delta$ definition.
The following is part of the proof to the above statement. 
proof. 
We have to show that given any $\epsilon>0$, there exists some $\delta>0$ such that $|f(x)-21|<\epsilon$ whenever $0<|x-3|<\delta$.
Working backwards:
$|f(x)-21|=|x^2+2x-15|=|(x+5)(x-3)|=|x+5||x-3|<|x+5|\delta$, 
if $|x-3|<\delta.$
We need an upper bound for $|x+5|$ when $x$ is close to $3$. If $|x-3|<1$, then $2<x<4$, so that $7<x+5<9$, so $|f(x)-21|=|x+5||x-3|<9|x-3|<9\delta$.
My question is why do we need an upper bound for $|x+5|$ when $x$ is close to $3$?
 A: You've shown that $|f(x)-21| < |x+5|\cdot \delta$ 
follows from $0<|x-3| < \delta$.  What is desired is showing that $|f(x)-21| $ can be made small (to be precise, that you can force $|f(x)-21| < \epsilon$ for any given $\epsilon > 0$) by making $\delta$ small.  The second term of the product $|x+5|\cdot \delta$ gets smaller as $\delta$ is diminished, so to show that the product as a whole gets smaller you need some way of ensuring that the left term $|x+5|$ doesn't increase in some manner that would make the product do anything other than get smaller.  In this case, from the triangle inequality $|x+5| \le |x-3| + 8 < \delta + 8$, you can show that $\delta \le 1$ implies $|x+5|<9$.
Note for instance if the product was instead something like $\left|\frac{1}{x-3}+5\right|\cdot \delta$, then the fact that the left hand term grows without bound as $x$ approaches 3 means that the product might not be forced to approach zero by making $\delta$ sufficiently small.
A: What we want is to prove that $|f(x)- 21| < \epsilon$ whenever $|x-3| < \delta = $ something derivable from epsilon.
So what that proof had found at that time was $|f(x) - 21| \le |x+5|\delta$ whenever $|x-3| < \delta$.  
So that will work if $\delta \le \frac {\epsilon}{|x+5|} $ as $|f(x)-21| < |x+5||x-3| < |x+5|\delta \le |x+5|*\frac {\epsilon}{|x+5|}= \epsilon$.
But if $|x+5|$ is unbounded then that's not a proper choice of $\delta$ as $\frac {\epsilon}{|x+5|}$ could be infinitely small.  In fact $\delta \le \frac {\epsilon}{|x+5|}$ would not be an acceptable choice for $\delta$ as it is not a constant and is dependent on $x$... The very thing we are trying to restrict!  We can't restrict a value based upon the value itself!
But by binding $|x + 5| < 9$ we have $|f(x) - 21| < |x+5||x-3| < |x+5|\delta < 9\delta$ and thus if $\delta \le \frac {\epsilon}{9}$ that is fine.  $\frac {\epsilon}9$ is a constant and if we set $\delta \le \frac {\epsilon}9$ is a perfectly acceptable value.
And whenever $|x-3| < \delta = \frac {\epsilon}9$.  We know $|f(x) - 21| < |x+5||x-3|< |x+5|\delta < 9 \delta \le 9*\frac {\epsilon}9 = \epsilon$.
