Doubt in epsilon-delta limit proof (more specifically, the inequalities) For example:
We shall prove that
$$ \lim_{x \to 2} x^2 = 4.$$
Then we shall prove the statement
$$ \forall \varepsilon &gt;0, \,\, \exists \delta &gt;0 \,\, \mid \,\, 0 &lt; \left| x - 2\right| &lt; \delta \implies \left|x^2 - 4 \right| &lt; \varepsilon$$

Draft:
$$ \left|x+2 \right| \left|x-2 \right| &lt; \varepsilon \qquad (1) $$
Let $\delta \leq 1$:
$$ \left|x - 2 \right| + |2| &lt; 1 + |2|$$
$$ |(x - 2) + (2)| \leq \left|x - 2 \right| + |2| $$
$$ |x| &lt; 1 + |2|$$
$$ x &lt; 3 $$
So
$$ |x-2||x+2| &lt; 5|x-2| \qquad (2)$$



And here comes the doubt!



$$ 5 \left|x - 2 \right| &lt; \varepsilon \qquad (3) $$



Proof:

Let
$$\delta = \min \left(1, \, \dfrac{\varepsilon}{5} \right)$$
[The proof continues, but I don't have trouble with the remainder part.]



Why can we say that $ 5 \left|x - 2 \right| &lt; \varepsilon $ ? If we add the inequalities (1) and (2), we won't get the inequality (3).
Can anyone help me?
Thank you.

Sorry for my bad English.
We know that $\lim\limits_{x\rightarrow2}{x}=2 \tag{1}$(Hint: Try $\delta=\varepsilon$)

Note that $x^2-4=\left(x-2\right)\left(x-2\right)+4(x-2)\tag{2}$

Using $(1)$ we can say that there exist a $\delta_1$ such that $|x-2|&lt;\sqrt{\frac{\varepsilon}{2}}$. Again using $1$ we can say that there exist a $\delta_2$ such that $|x-2|&lt;\frac{\varepsilon}8$.

Now,there exist a $\delta=\min\{\delta_1,\delta_2\}$ such that $(2)$(with the help of triangular inequality) can be reduced to $$\left|x^2-4\right|&lt;\left|\left(x-2\right)\right|\left|\left(x-2\right)\right|+\left|4(x-2)\right|&lt;\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
There are several issues going on here.  To answer your first question, we can't conclude $5|x-2| &lt; \epsilon$.   

Pretend I have chosen my favorite small number $\epsilon$.  To find $\delta$, I need to figure out how close $x$ must be to $2$ in order to make $|x^2 -4| &lt; \epsilon$.  So from this, hopefully you can see that $\delta$ depends on $\epsilon$.  The smaller $\epsilon$ is, the smaller $\delta$ will need to be.  Therefore, you should not have the line "Let $\delta \leq 1$:" in your proof.  Rather, you should have 

"Let $\delta = $ (some expression involving $\epsilon$)"

So maybe try something like "Let $\delta = \sqrt{\epsilon}$" and see what you can come up with.  (Note: this is not a correct value of $\delta$, it's just an example to get you going.)