I have to prove that $\lim\limits_{x \rightarrow a} 7x^2$ exists for every $a \in \mathbb{R}$ by using an $\epsilon - \delta$ proof. I started by making a proof strategy:
To prove: $\lim\limits_{x \rightarrow a} 7x^2 = 7a^2$.
Let $\epsilon >0$. What we are required to do is to find a number $\delta >0$ such that if $0 < |x-a| < \delta$, then $|7x^2 - 7a^2| < \epsilon$. From $|7x^2 - 7a^2|$ follows:
$|7x^2 - 7a^2| = |7x + 7a||x-a| < \epsilon$. We seek an upper bound for $|7x + 7a|$. To do this, we impose the restriction $\delta \leq 1$. Thus $|x-a| < \delta \leq 1$. So $-1 < x-a < 1$ and $-1 + 2a < x+a < 1 +2a$. Thus $-7 + 14a < 7x + 7a < 7 +14a$. Hence,
$|7x + 7a| < 7 +14a$.
From here follows $|7x^2 - 7a^2| = |7x + 7a||x-a| < 7+14a |x-a| < \epsilon$, so $|x-a| < \dfrac{\epsilon}{7+14a}$. So this is why I chose $\delta = min(1, \dfrac{\epsilon}{7+14a})$, but of course this doesn't work for all real numbers $a$, since $\delta$ becomes negative whenever $a< - \frac{1}{2}$. Also, whenever $a = - \frac{1}{2}$, $\dfrac{\epsilon}{7+14a}$ is not defined. So I am doing something wrong, but I am not sure how I can find my right $\delta$.
Edit: I might have found the solution to solve the problem of $\delta$ being negative. I just choose $\delta = min(1, \dfrac{\epsilon}{|7 +14a|})$, since $7+14a \leq |7+14a|$, so if $|7x + 7a| < 7+14a$, then also $|7x + 7a| < |7+14a|$. But then again, I don't know how to solve the problem of $\dfrac{\epsilon}{|7 +14a|}$ not being defined when $a = - \frac{1}{2}$.
If $0 < |x-a| < \delta$ then the reverse triangle inequality gives $$|x| - |a|\le |x-a| < \delta \implies |x| < |a|+\delta$$
so $$|7x+7a| \le 7(|x|+|a|) \le 7(2|a|+\delta)$$ If $\delta \le 1$ then we get $|7x+7a|\le7(2|a|+1) = 7+14|a|$ which is always positive.
Hence for $\delta \le \min\left\{1, \frac{\varepsilon}{7+14|a|}\right\}$ we have $$|7x^2-7a^2| \le |7x+7a||x-a| < (7+14|a|)\delta \le \varepsilon$$