Epsilon delta proof verification

Question

Prove that $ \lim_{x \to a} 5x^3$exists for every $a \in \mathbb{R}$.

Here's my proof. I was wondering if it is complete and notationally correct:

Suppose $\epsilon > 0$ has been provided. Let $\epsilon _2 = \min\{ \epsilon, 4a^3\}$. We define $\delta = \min \{ -a + \sqrt[3]{a^3-\dfrac{\epsilon}{5}}$, $-a + \sqrt[3]{a^3+\dfrac{\epsilon}{5}} \}$. Since $\epsilon _2 > 0$, we also have $\delta > 0$.

$$ -a + \sqrt[3]{a^3-\dfrac{\epsilon}{5}} < x-a < -a + \sqrt[3]{a^3+\dfrac{\epsilon}{5}}$$

$$5a^3 - \epsilon _2 < 5x^3 < 5a^3 + \epsilon_2$$

$$|5x^3-5a^3| < \epsilon_2 < \epsilon $$

Hence, $ \lim_{x \to a} 5x^3 = 5a^3$ for every $a \in \mathbb{R}$. Therefore, the limit exists for every $a \in \mathbb{R}$.

So this is the entirety of my proof. Is it correct?

Answer 1

I would say no, what about if both of the possible values of $\delta$ are negative? You have no value you can use.

The easiest proof would be taking the definition: Given $\epsilon > 0, $ exists $\delta>0$ such that: $$|x-a|<\delta \implies |5x^3-5a^3|<\epsilon$$

First you could assume $\delta < 1$, so: $$|x-a|<\delta \implies x < a+1$$ Then $|5x^3-5a^3| \leq 5|x-a||x^2-xa+a^2| \leq 5|x-a|(|x^2|+|x||a|+|a^2|)$

$$5|x-a|(|x^2|+|x||a|+|a^2|) \leq 5|x-a|(|a+1|^2+|a+1||a|+|a|^2) < \epsilon$$

Then, you can take $\displaystyle\delta = min(1,\frac{\epsilon}{5(|a+1|^2+|a+1||a|+|a|^2)})$, then you can do the process backwards and conclude that the definition of limit is true given that $\delta$ value.

Answer 2

If you know the following difinition of continuous function, this will helpful.

If $f(x)$ is continuous at x=a iff that $\lim_{x\to a}f(x)=f(a)$.