# Thread: Inequalities concerning limits

1. ## Inequalities concerning limits

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 $f$ defined in some open interval containing $a$ (but not necessarily at $a$ itself), we say
$\lim_{x \to a}f(x)=L$,
if given any number $\epsilon > 0$, there is another number $\delta > 0$, such that $0 < \left\vert x-a \right\vert < \delta$ guarantees that $\left\vert f(x)-L \right\vert < \epsilon$.

Now this is shown in an example where it is to be proved that $\lim_{x \to 2}x^2=4$.
Analagous to the definition it says there must be a $\delta > 0$ for which $0 < \left\vert x-2 \right\vert < \delta$ guarantees that $\left\vert x^2-4 \right\vert < \epsilon$ for any given $\epsilon$.

Next thing it says is the following:
$\left\vert x^2-4 \right\vert = \left\vert x+2 \right\vert \left\vert x-2 \right\vert$,
which makes sense since it's just factoring.

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

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

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

Greetings

2. ## Re: Inequalities concerning limits

Originally Posted by Floele1106
Now my actual problem:
Why is (and that's what the textbook says) $\left\vert x^2-4 \right\vert \le 5\left\vert x-2 \right \vert$?
We multiply both sides of the inequality |x + 2| ≤ 5 by |x - 2|. Since |x - 2| is nonnegative, the direction of the inequality does not change.

3. ## Re: Inequalities concerning limits

Well, that definitely makes sense. Thanks a lot.