Proof limit by definition Prove: $\lim_{x \to 2}x^2=4$


We need to find $\delta&gt;0$ such that for all $\epsilon&gt;0$ such that if $|x-2|&lt;\delta$:

$$|x^2-4|=|(x-2)(x+2)|=|x+2|\cdot|x-2|\implies|x-2|&lt;\frac{\epsilon}{|x+2|}$$

We will take $\delta=\frac{\epsilon}{|x+2|}$

is it valid?
If $x \in \Bbb{R}$, then
$$
|x^{2}-4| = |x-2||x+2|;
$$
if in addition $|x-2| &lt; 1$, then $|x| - 2 \leq |x-2| &lt; 1$, implying $|x| &lt; 3$, implying $|x+2| \leq |x|+2 &lt; 5$, implying 
$|x-2||x+2| &lt; 5|x-2|$;
given any $\varepsilon &gt; 0$, we have $5|x-2| &lt; \varepsilon$ if in addition $|x-2| &lt; \varepsilon/5$. In conclusion, for every $\varepsilon &gt; 0$, if $|x-2| &lt; \min \{ 1, \varepsilon/5 \}$ then $|x^{2}-4| &lt; \varepsilon$.

Note that the choice of $\delta$ you made is not gonna work, for it depends on $x$; recall the definition of limit.
You can cheat a little bit, as long as it doesn't appear in your final solution: At $x = 2$, the derivative of $x^2$ is $4$. That means that $\delta$ must be, for small $\epsilon$, less than $\frac\epsilon4$.

For small $\delta$, $x$ is close to $2$, which makes $|x+2|$ close to $4$. That means you're on the right track, but you have to eliminate the $x$ from that expression in some way, probably by saying something like $\delta = \min(\frac14, \frac\epsilon5)$. (I am quite certain that this is good enough, by the way, but you would have to check it; I only went with what felt right here.)