Why is this step necessary to prove $\lim_{x \to 2} x^2 = 4$? I'm looking at third exercise in this page.
In the end of step 3, why is it necessary to assume that
\begin{equation}
5|x - 2| < \varepsilon
\end{equation}
Why would it be wrong if I had just stated that
\begin{align*}
|x + 2||x - 2| &< 5|x - 2|\\
&< 5\delta
\end{align*}
And if $5\delta = \varepsilon$, then the first implication is true when $|x - 2| < 1$. Thus, we can choose $\delta = \min\{1, \frac{\varepsilon}{5}\}$, because this would ensure $|x - 2| < 1$.
 A: First, the author does not assume that! The author only uses a phrase like "we want $\cdots$". 
Let me show you a way to prove the proposition in one stroke. If $x \in \mathbb{R}$, then $|x^{2} - 4| = |x-2||x+2|$. The term $|x+2|$ is annoying here, so we may bound it away by bounding $|x-2|$. If in addition $|x-2| < 1$, then $|x| - 2 \leq |x-2| < 1$ by triangle inequality; so $|x| < 3$, and hence $|x+2| \leq |x| + 2 < 5$ by triangle inequality again. Then $|x-2||x+2| < 5|x-2|$. (note that we are still under the assumption that $|x-2| < 1$.) Given any $\varepsilon > 0$, we have $5|x-2| < \varepsilon$ if in addition $|x-2| < \varepsilon/5$ (corresponding to the author's "we want $\cdots$".). Now putting all these together suggests that taking $\delta := \min \{ 1, \varepsilon/5 \}$ suffices. 
To be honest, your argument in its present form seems to me not convincing; it seems "fractional" you know. The most suspicious part is the paragraph beginning with "And". You may want to take some time to rethink how to phrase your thought. Perhaps you know it in your head, but it is just that you have not expressed it accurately.
