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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?

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  • $\begingroup$ Everything is right from the double-inequality onward, but your definition of $\delta$ is a bit off: the second value works for the upper limit, but the first value is negative of what you want. Even once you have this, though, you still would need to show that that implies the inequality; it is easy to imagine, if you don't do any checking, that $a-(a^3-\varepsilon/5)^{1/3}<-a+(a^3+\varepsilon/5)^{1/3}$. $\endgroup$ Oct 21, 2014 at 3:44

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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.

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  • $\begingroup$ How can you assume that delta is less than one? Why does that work? $\endgroup$
    – HiiiPower
    Oct 21, 2014 at 1:46
  • $\begingroup$ @HiiiPower: Remember, your job is that, if you are given a random $\varepsilon$, you must find a $\delta$ such that the constraints are satisfied. Choosing a smaller $\delta$ will satisfy these constraints simply restricts the possible values of $x$ which can be plugged into the $\varepsilon$ inequality, and therefore if $\delta_1<\delta_2$ and $\delta_2$ satisfies the constraints, then so does $\delta_1$. $\endgroup$ Oct 21, 2014 at 1:50
  • $\begingroup$ I can assume delta is less than one because I don't have a fixed value of delta, but I know that if delta is lesser than 1, I have that chance to make the transformation I made. So I take delta the minimun between 1 and that stuff, and so it works because, if the fraction is lesser than 1, then everything is true. If the fraction is greater than 1, everything is true again. $\endgroup$
    – Rono
    Oct 21, 2014 at 1:51
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If you know the following difinition of continuous function, this will helpful.

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

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