Do I have to specify $x \in I$ for this limit? Let $a > 0$ and $I = (0,a)$. I want to prove using the definition of limit that $$\lim_{x\to c} x^2 = c^2$$ for all $c \in I$. 
Let $\varepsilon > 0$ be given. Since this inequality $|x^2 - c^2| \leq 2a|x - c|$ holds for all $x,c\in I$, I can take $\delta = \varepsilon/(2a)$ as long as $x\in I$. Can I just say that $x \in I$ or  is it better to choose $$\delta = \inf \{c, a - c, \varepsilon/(2a)\}?$$
I have solve the more general problem (that is for all $c\in \mathbb{R}$) with another approach. Just curious for this one.
 A: You need to select a value for $\delta$ which satisfy the definition. 
I think that $\delta = \frac{\varepsilon}{2a}$ is not a good choice since, for example, for $a=\frac12$ we obtain $\delta = \varepsilon$  which does not satisfy the definition. 
Indeed suppose $c=\frac12$ and $x=\frac12+k\varepsilon$ where $k\in(0,1)$, thus:
$$x^2=\frac14+k^2\varepsilon^2+k\varepsilon\implies x^2-\frac14=k^2\varepsilon^2+k\varepsilon=\varepsilon(k^2\varepsilon+k)>1 \quad \forall k\in(k_0,1)$$
The value of $k_0$ can be easily found by the condition
$$k^2\varepsilon+k>1\implies k^2\varepsilon+k-1>0\implies k>k_0=\frac{\sqrt{1+4\varepsilon}-1}{2\varepsilon} $$
plot of $k_0=\frac{\sqrt{1+4\varepsilon}-1}{2\varepsilon}$
You can proceed as follow. 
We need to show that
$$\forall \varepsilon > 0 \quad \exists\delta>0 \quad0<|x-c|<\delta\implies|x^2-c^2|<\varepsilon$$
Let consider
$$|x^2-c^2|<\varepsilon\iff-\varepsilon < x^2-c^2< \varepsilon\iff c^2-\varepsilon < x^2< c^2+\varepsilon$$
Case 1: $c^2-\varepsilon\leq0$
$$c^2-\varepsilon < x^2< c^2+\varepsilon\iff0<x<\sqrt{c^2+\varepsilon}\iff-c<x-c<\sqrt{c^2+\varepsilon}-c$$
$$\iff|x-c|<\delta=\min\{|c|,\sqrt{c^2+\varepsilon}-c\}$$
Case 2: $c^2-\varepsilon>0$
$$c^2-\varepsilon < x^2< c^2+\varepsilon\iff\sqrt{c^2-\varepsilon}<x<\sqrt{c^2+\varepsilon}$$
$$\iff\sqrt{c^2-\varepsilon}-c<x-c<\sqrt{c^2+\varepsilon}-c$$
$$\iff|x-c|<\delta=\min\{c-\sqrt{c^2-\varepsilon},\sqrt{c^2+\varepsilon}-c\}=\sqrt{c^2+\varepsilon}-c$$
A: It's probably better to choose the delta you gave based on $\inf$ (or, equivalently, $\min$, since it's over a finite set), and then show that that implies $x \in I$. Since the domain of the function (seemingly) is $\mathbb{R}$, not $I$, nothing in the definition of the limit guarantees $x \in I$.
(Note that if you had a closed interval $I = [0,a]$, your approach would not work, even though your inequality still holds for all $x,c \in I$! So I think you do have to justify assuming $x \in I$.)
