<s>[INST] limits calculus I am having trouble understanding part of the solution to this simple problem.
$\lim_{x \to 2} (x^2 + 3x) = 10$
Solution:
Let $\epsilon > 0$
$| x - 2 | < \delta$  and $| x^2 +3x -10 | < \epsilon$
since $x^2 +3x -10 = (x - 2)^2 + 7x -14 = (x - 2)^2 + 7x -14 = ( x -2 )^2 +7(x-2)$
$|(x-2)^2 +7(x-2)| \leq |(x-2)|^2 +7|(x-2)|$
$\delta^2 + 7\delta < \epsilon$
let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.
then $8\delta < \epsilon$
$\delta < \epsilon/8$.
My Question:
I worked my way through the question down to $\delta^2 + 7\delta < \epsilon$
I then got confused by the end of this statement 
Let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.
and in particular $\delta^2 \leq \delta$. I see how this allows me to prove the limit but I cannot make sense out of $\delta^2 \leq \delta$.
Could anyone explain this to me?
 [/INST] We have been challenged with an $\epsilon$, perhaps $\epsilon=1/1000$.  We want to come up with a $\delta$ such that if $|x-2|\lt \delta$, then for sure $|x^2+3x-10|\lt \epsilon$.
Suppose that after some calculation, we announce that $\delta=\epsilon/8$ does the job. Then triumphantly the challenger could say that she had $\epsilon=47$ in mind, and that in that case taking $\delta=47/8$ is insufficient. Of course, that is not playing fair. But we might as well come up with a $\delta=\delta(\epsilon)$ that always works.  
Some algebra shows that $x^2+3x-10=(x-2)^2+7(x-2)$. So we want to make sure that $|(x-2)^2+7(x-2)|\lt \epsilon$. Note that 
$$|(x-2)^2+7(x-2)|\le (x-2)^2+7|x-2|.$$
We want to make the right-hand side "small," by choosing $x$ appropriately close to $2$. Suppose we had been given a ridiculous $\epsilon$, like $47$. If $x$ is within $47$ of $2$, the number $(x-2)^2$ could be very large. So the first task is to make sure $\delta$ is small enough not to allow $(x-2)^2$ to be large.  
So we say first of all, let $\delta\le 1$. Then if $|x-2|\lt \delta$, it follows that $(x-2)^2\lt \delta$.  For $(x-2)^2=(x-2)(x-2)$. The "first" $x-2$ has absolute value $\le 1$, and the second has absolute value $|x-2|$, so the product has absolute value $\le |x-2|$.
It follows that as long as $\delta\le 1$, we have $(x-2)^2+7|x-2|\le 8|x-2|$. To make sure this is $\lt \epsilon$, it is enough to make $\delta=\frac{1}{8}\epsilon$.
However, in deriving our simplified inequality, we assumed that $\delta\le 1$. So we know that everything will work if $\delta=\min(1,\epsilon/8)$.  
It might have been better to observe that
$$(x-2)^2+7|x-2|=|x-2|\left(|x-2|+7\right).$$
Then it is clear that if $|x-2|\le 1$, then $(x-2)^2+7|x-2|\le 8|x-2|$. 
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