Hi you guys,

so I have some difficulty understanding some of the content in my textbook. I recently started learning about limits and now I got to the formal defintion of a limit in my textbook which sounds/looks perfectly reasonable to me:

For a function $\displaystyle f$ defined in some open interval containing $\displaystyle a$ (but not necessarily at $\displaystyle a$ itself), we say

$\displaystyle \lim_{x \to a}f(x)=L$,

if given any number $\displaystyle \epsilon > 0$, there is another number $\displaystyle \delta > 0$, such that $\displaystyle 0 < \left\vert x-a \right\vert < \delta$ guarantees that $\displaystyle \left\vert f(x)-L \right\vert < \epsilon$.

Now this is shown in an example where it is to be proved that $\displaystyle \lim_{x \to 2}x^2=4$.

Analagous to the definition it says there must be a $\displaystyle \delta > 0$ for which $\displaystyle 0 < \left\vert x-2 \right\vert < \delta$ guarantees that $\displaystyle \left\vert x^2-4 \right\vert < \epsilon$ for any given $\displaystyle \epsilon$.

Next thing it says is the following:

$\displaystyle \left\vert x^2-4 \right\vert = \left\vert x+2 \right\vert \left\vert x-2 \right\vert$,

which makes sense since it's just factoring.

Because our sole concern is what's in close proximity to $\displaystyle x$, it is assumed that $\displaystyle x$ be in the interval $\displaystyle \left [ 1,3 \right ]$ and from that it follows that $\displaystyle \left\vert x+2 \right\vert \le 5$.

Now my actual problem:

Why is (and that's what the textbook says) $\displaystyle \left\vert x^2-4 \right\vert \le 5\left\vert x-2 \right \vert$?

That's probably some simple algebra and I'm overlooking something, but I really appreciate your help and thank you in advance.

Greetings