Question: Define $p(x) = 2x^3-7x$. Give an $\epsilon - \delta$ proof that $\lim_{x \to a} p(x)$ exists for every $a \in R$.
My attempt: Claim that the $\lim_{x \to a} 2x^3-7x = 2a^3 - 7a$.
Given an arbitrary $\epsilon > 0$, $\exists\delta>0$ such that $0<|x-a|<\delta \implies |f(x)-L|>\epsilon$.
(i.e) $0<|x-a|<\delta \implies |2x^3-7x-(2a^3-7a)|<\epsilon$
Consider,
$$|2x^3-7x-(2a^3-7a)| = |2x^3-2a^3+7a-7x|$$ $$ \le |2||x^3-a^3|+|7||x-a|$$ by triangle inequality.
So,
$$|x-a|< \frac{\epsilon - |2||x^3-a^3|}{|7|}$$
Because $|x^3-a^3|$ is a problem point, I need to control it, so the way I tried to do that is as follows:
If $|x-a|<1$ then $$|x-a|<1 \implies |x^3-a^3|<|a^2+ax+x^2| (*)$$
Here's where I'm stuck. I'm not sure what to do with this: $|a^2+ax+x^2|$. Usually for proof like this I end up with something like $\delta = \min\left\{1, \frac{\epsilon-2(*)}{7}\right\}$ but in this case the result from $(*)$ wasn't in some form of a number, so I am not sure what to do next. If someone can help me with this that would be really appreciated.
Suggestion: think about the aim of the problem. You are going to say $|x-a|<\delta$, for some suitable $\delta$; to make full use of this you need to get everything possible in terms of $|x-a|$, including the $x^2+ax+a^2$.
If $|x-a|<1$ then $$\eqalign{|x^2+ax+a^2| &=|(x-a)^2+3a(x-a)+3a^2|\cr &\le|x-a|^2+3|a||x-a|+3|a|^2\cr &<1+3|a|+3|a|^2\cr}$$ which gives $$|(2x^3-7x)-(2a^3-7a)|<(9+6|a|+6|a|^2)|x-a|\ .$$