2

Prove $\lim_{x \to 0} x^3+x^2 = 0$ using delta-epsilon.

Choose $\delta < min \{(\frac{\epsilon}{2})^{1/3}, (\frac{\epsilon}{2})^{1/2} \}$

Assume $0 <|x| < \delta$.

Therefore $|x|^3 < \frac{\epsilon}{2}$ and $|x|^2 < \frac{\epsilon}{2}$. So $|x^3+x^2| \leq |x|^3 + |x|^2 < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Something tells me there's an error here, because it appears that we need delta to be less than 1. But I can't come up with a counterexample to invalidate my proof.

Snowball
  • 1,089
  • What suggests $\delta<1$? – Jam Nov 16 '19 at 16:27
  • @Jam A cryptic comment in the notes. I have a hint as to how to start, and it started with picking $\delta <1$. At the end there was a cryptic question at the end after giving this hint. Can we choose $\delta$ such that $\delta^3 < \epsilon/2 $ and $\delta^2 < \epsilon/2$, as if to suggest there is an error in doing it like this. – Snowball Nov 16 '19 at 16:29
  • I'm not sure but I've got a feeling that was meant to refer to the fact that it's only past $\varepsilon=2$, that the inequality $(\varepsilon/2)^{1/2}<(\varepsilon/2)^{1/3}$ reverses. This is also the point where both functions of $\varepsilon$ equal $1$. So simply assuming $\delta<1$ gives us that $\min{(\varepsilon/2)^{1/2},(\varepsilon/2)^{1/3}}=(\varepsilon/2)^{1/2}$ and you could do the rest of the proof ignoring the $\min$. I don't think there's anything wrong with your way though. – Jam Nov 16 '19 at 16:43

1 Answers1

1

You proof is correct however you could simplify your life by considering $|x|<1$.

$|x|<1\implies |x|^2<|x|\implies |x|^3<|x|^2$

Thus $|x^3+x^2|<|x|^3+|x|^2<2|x|\to 0$ when $x\to 0$

This is why taking $\delta=\min(1,\varepsilon)$ is helping.

Another reason you were expecting $\delta<1$ is that it is often used in epsilon-delta proofs.

We often consider $\delta=\min(1,\varepsilon)$

  • The condition $\delta<\varepsilon$ is used to show that some quantities can be made arbitrary small.

  • The condition $\delta<1$ is used to show that some other quantities are bounded.

You generally end up with $|f(x)-\ell|<K\varepsilon$ where $K$ is a constant, which is sufficient to conclude.

( i.e. you can take $\delta'=\frac 1K\delta$ and get $|f(x)-\ell|<\varepsilon$, but this step is not absolutely required ).

Here is such an example: https://math.stackexchange.com/a/2486840/399263

In the present case, we could do instead $\quad x^3+x^2=x^2(1+x)$

  • it is split into the arbitrary small quantity $|x|^2<\varepsilon$ for $\delta<\sqrt{\varepsilon}$

  • and the bounded part $|1+x|<2$ whenever $|x|<1$

Thus taking $\delta=\min(1,\sqrt{\varepsilon/2})\implies |x^3+x^2|<\varepsilon$

As you can see there are many ways of dealing with the limit at hand.

However simple inequalities like $|x|<1\implies |a_nx^n+\cdots+a_1x|<|x|\sum|a_i|$ are often ignored althought simplifying considerably the calculations.

Always consider simplifying an expression with raw inequalities before going for details about epsilon-delta proofs.

zwim
  • 28,563