Delta Epsilon Proof Check: Show $\lim_{x \to 0} x^3+x^2 = 0$ 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.
 A: 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.
