Use the definition of a limit to prove that $\displaystyle\lim_{x \to 5}x^2 = 25$ $\displaystyle\lim_{x \to 5}x^2 = 25$
Attempt:
We want to show that $\forall \epsilon > 0, \exists \delta > 0$ such that if $0 < |x - 5| < \delta$, then $|x^2 - 25| < \epsilon$.
$|x^2 - 25| = |(x+5)(x-5)| = |x+5||x-5|$. Here we have $|x-5| < \delta$, but now we need $|x+5| < \delta$. 
Assume $\delta = 1$. Then $|x-5| < 1 \Rightarrow -1 < x-5 < 1 \Rightarrow 4 < x < 6$. 
From here we let $\delta = \min\{\displaystyle\frac{\epsilon}{11}, 1 \}$
Then we have that $|x^2 - 25| < 11|x-5|< 11 \displaystyle\frac{\epsilon}{11} = \epsilon$.
If my proof is unorganized, out of order, or missing something I would appreciate the feedback. I do get uncomfortable with epsilon delta proofs so I wanted to be sure.
 A: So, the above is filled with how you derived information.
A better proof is more direct:

Given $\epsilon>0$, let $\delta=\min\left(1,\frac{\epsilon}{11}\right)$.
Step one: Show that when $|x-5|<\delta$ then $|x+5|<11$.
Step two: Use step one to show that when $|x-5|<\delta$ then $$|x^2-25|=|x+5||x-5|< 11\cdot \frac{\epsilon}{11}=\epsilon.$$

Aside: Your claim that you need $|x+5|<\delta$ is completely wrong. If $|x-5|<\delta$ then $|x+5|>10-\delta$. So, unless $\delta>5$, that's not going to be true often.
What you needed what $\delta|x+5|<\epsilon$.
A: By what you went from $4 < x < 6$ to the choice $\varepsilon/11$ is unclear; let me provide an argument for your reference:
If $x \in \Bbb{R}$, then $|x^{2}-25| = |x-5||x+5|$; if in addition $|x-5| < 1$, then $|x| - 5 \leq |x-5| < 1$, implying $|x| < 6$, implying $|x+5| \leq |x| + 5 < 11$, implying $|x-5||x+5| < 11|x-5|$; given any $\varepsilon > 0$, we have $11|x-5| < \varepsilon$ if in addition $|x-5| < \varepsilon/11$. Hence we conclude that, for every $\varepsilon > 0$, we having $|x-5| < \min \{1, \varepsilon/11 \}$ implies $|x^{2}-25| < \varepsilon$.
A: I think you should write it out as follows:
Let $\epsilon >0$. Notice that if $\left|x-5\right|<1$, then $4<x<6$ and so $\left|x+5\right|=x+5<11$. Thus let $\delta=\min \{\epsilon/11,1\}$ and assume $0<\left|x-5\right|<\delta$. Then both $\left|x-5\right|<1$ and $\left|x-5\right|<\epsilon/11$. So $\left|x^2-25\right|=\left|x-5\right|\left|x+5\right|<\left|x-5\right|\cdot 11<(\epsilon/11)\cdot11=\epsilon$. Therefore, $\lim_{x\rightarrow5}x^{2}=25$.
A: we have $|x-5| \le \delta$ so $5-\delta \le x \le 5+\delta$ which implies $|(x-5)(x+5)| \le \delta(10+\delta)\le \epsilon$  
To fixe $\delta$, it is sufficient to solve the second degree inequation in $\delta$
$\delta^2+ 10\delta -\epsilon \le 0$ with discriminant $\Delta= 100 +4\epsilon >0$
and to get $\delta$ between the two solutions $\delta_1$, $\delta_2$ of $\delta^2+ 10\delta -\epsilon = 0$.
So, for each $\epsilon>0$, there is "a" (in fact, an infinity) $\delta$ ...
