How to find the $\delta$ for some $\epsilon$ such that: $|x−1|<\delta\implies |f(x)−f(1)|<\epsilon$

Question

$f(x)$ is defined as:
$f(x)=\frac{\sqrt[3]{x}-1}{x-1};x\in \mathbb{R} \setminus \{ 1 \} $
$f(x)=\frac{1}{3};x=1$

How to find the $\delta$ for some $\epsilon$ such that: $|x−1|<\delta\implies |f(x)−f(1)|<\epsilon$?

I tried a lot of things, but I seem to be completely stuck.

Answer 1

Note that $$\frac{\sqrt[3]{x}-1}{x-1} = \frac{\sqrt[3]{x}-1}{x-1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1} = \frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}.$$

This follows from the equality $a^3 - b^3 = (a -b)(a^2 +ab+b^3)$.

Answer 2

We'll be using this simple inequality, true for all $x, y \geq 0$: $$ \sqrt[3]{x + y} \leq \sqrt[3]{x} + \sqrt[3]{y} \text{.}$$ Indeed, it's equivalent to $$ x + y \leq x + 3 \sqrt[3]{x^2 y} + 3 \sqrt[3]{x y^2} + y \text{,}$$ which clearly holds when $x, y \geq 0$.

Now, we want to prove that $$\forall_{\varepsilon > 0} \exists_{\delta > 0} \ |x-1| < \delta \Rightarrow \left| \frac{\sqrt[3]{x}-1}{x-1} - \frac{1}{3} \right| < \varepsilon.$$ As it's been said, that's the same as $$\forall_{\varepsilon > 0} \exists_{\delta > 0} \ |x-1| < \delta \Rightarrow \left| \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1} - \frac{1}{3} \right| < \varepsilon.$$ Let's set our $\varepsilon > 0$. First, if $\varepsilon > 1$, we can just take $\delta = 1$, since then: $$ -\varepsilon < 0 -\frac{1}{3} < \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1} - \frac{1}{3} < \frac{1}{0 + 0 + 1} - \frac{1}{3} < \varepsilon.$$ So, from now on, we can assume $\varepsilon \leq 1$ (and we will often use that). We'd like to show now that it will suffice if for our chosen $\delta$, $$ \left| \sqrt[3]{x^2} + \sqrt[3]{x} -2 \right| < 2\varepsilon. $$ This will let us focus on the real challenge of the task. Let's put $d := \sqrt[3]{x^2} + \sqrt[3]{x} - 2$. We have: $$ \left| \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1} - \frac{1}{3} \right| = \left| \frac{1}{2 + \left( \sqrt[3]{x^2} + \sqrt[3]{x} - 2 \right) + 1} - \frac{1}{3} \right| = \left| \frac{1}{3 + d} - \frac{1}{3} \right| = \left| \frac{-d}{9 + 3d} \right|.$$ Then, assuming $|d| < 2\varepsilon$, we get $$ \left| \frac{-d}{9 + 3d} \right| = \frac{|d|}{\left| 9 + 3d \right|} \leq \frac{|d|}{\left| |9| - |3d| \right|} = \frac{|d|}{9 - |3d|} \leq \frac{|d|}{9-6} \leq \frac{2\varepsilon}{3},$$ which is what we desired.

Now we just need to pick a $\delta$, such that if $\left| x - 1 \right| < \delta$, then $ \left| \sqrt[3]{x^2} + \sqrt[3]{x} -2 \right| < 2\varepsilon$. Again, let's put $y := x - 1$ for clarity of notation. First, we'll consider the case when $y \geq 0$ and will find how small a $\delta$ we need to take care of that case. Then we'll attack the less convenient case when $y < 0$. Assuming $y \geq 0$ we have $$ \left| \sqrt[3]{x^2} + \sqrt[3]{x} -2 \right| < 2\varepsilon \Leftrightarrow \left| \sqrt[3]{{(1+y)}^2} + \sqrt[3]{1+y} -2 \right| < 2\varepsilon \Leftrightarrow \sqrt[3]{{(1+y)}^2} + \sqrt[3]{1+y} -2 < 2\varepsilon. $$ We can replace this inequality with a stronger one - then if our $\delta$ works for the stronger version, it will definitely work for the former one too. Let's do it using the inequality from the beginning. $$ \sqrt[3]{{(1+y)}^2} + \sqrt[3]{1+y} -2 < 2\varepsilon \Leftarrow \sqrt[3]{1} + \sqrt[3]{2y + y^2} + \sqrt[3]{1} + \sqrt[3]{y} - 2 < 2\varepsilon \Leftrightarrow \sqrt[3]{2y + y^2} + \sqrt[3]{y} < 2\varepsilon.$$ Now, since we'll be picking $\delta \leq 1$, which will cause $y \leq 1$, we can assume just that. Then $y^2 \leq y$ and we can continue: $$ \sqrt[3]{2y + y^2} + \sqrt[3]{y} < 2\varepsilon \Leftarrow \sqrt[3]{3y} + \sqrt[3]{y} < 2\varepsilon \Leftrightarrow \left( \sqrt[3]{3} + 1 \right) \sqrt[3]{y} < 2\varepsilon \Leftarrow 4\sqrt[3]{y} < 2\varepsilon.$$ So it will be enough if $\delta = {\left( \frac{\varepsilon}{2} \right)}^3.$

Now for the case when $y < 0$. Then our inequality is of the form $$ 2 - \sqrt[3]{1+2y+y^2} - \sqrt[3]{1+y} < 2\varepsilon. $$ We do something similar as before, just remembering $y < 0$: $$ 2 - \sqrt[3]{1+2y+y^2} - \sqrt[3]{1+y} < 2\varepsilon \Leftarrow 2 - \sqrt[3]{1+2y} - \sqrt[3]{1+y} < 2\varepsilon \Leftarrow 2 - 2\sqrt[3]{1+2y} < 2\varepsilon.$$ Now we can cube both sides to find an appropriate $\delta$. $$ 2 - 2\sqrt[3]{1+2y} < 2\varepsilon \Leftrightarrow \sqrt[3]{1+2y} > 1 - \varepsilon \Leftrightarrow 1+2y > 1 - 3\varepsilon + 3{\varepsilon}^2 - {\varepsilon}^3 \Leftrightarrow 2y > - 3\varepsilon + 3{\varepsilon}^2 - {\varepsilon}^3.$$ Now we just strengthen the inequality a bit more (again using $\varepsilon \leq 1$) and find what we want: $$ 2y > - 3\varepsilon + 3{\varepsilon}^2 - {\varepsilon}^3 \Leftarrow 2y > - 3\varepsilon + 3{\varepsilon} - {\varepsilon}^3 \Leftrightarrow 2y > - {\varepsilon}^3 \Leftrightarrow |y| < \frac{{\varepsilon}^3}{2}. $$ So it will be enough for this case if $\delta = \frac{{\varepsilon}^3}{2}$.

In the end, we can take $$ \delta = \min \left\{ 1, \left( \frac{\varepsilon}{2} \right)^3, \frac{\varepsilon ^ 3}{2} \right\}, \quad \text{which reduces to} \quad \delta = \min \left\{ 1, \frac{\varepsilon ^ 3}{8} \right\}.$$ This is clearly greater than $0$, so the thesis holds.

This has turned out to be absurdly long, but I've already written it all, so here you have it :)

Keep in mind, there will probably be a mistake somewhere here, it's just too long. Hopefully though, it won't make the proof completely wrong. I'm also sure it can be done more simply, but I don't know how much. It seems to me that whoever gave you this task, they weren't in their best mood at that time.