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?