Can I assume that $x + x^2 < 1$ if $x \to 0$?

Question

I'm trying to solve a limit, and got to this, [] being the floor function:

$$ \lim_{x\to0^+} \frac{[x+x^2]}{x} $$

I know that $$\lim_{x\to0^+} x + x^2 = 0$$

So I want to say that the $[x+x^2] = 0$ (not tends to 0, but be equal).

For that I'd need to prove that $ 0 < x + x^2 < 1$

Is it rigorous enough to just say that since the limit tends to $0$ then the floor evaluates exactly to $0$? This would mean the limit is:

$$ \lim_{x\to0^+} \frac{[x+x^2]}{x} = \lim_{x\to0^+} \frac{0}{x} = 0 $$

Which I know to be truth, I'm just wondering if the reasoning is rigorous enough. It basically comes down to being able to assume that anything that tends to 0 is arbitrarily small, without losing generality or anything like that. Can I do it?

Answer 1

You can write $$x^2+x =x(x+1) \approx x(1)=x \rightarrow 0^+$$

Since : $$x^2+x \rightarrow 0^+ \implies \lfloor x^2+x \rfloor = 0 $$

Answer 2

If you have proven that the numerator is identically $0$ in an interval $[0,\epsilon]$ with $\epsilon >0$ (which works here), you have actually shown that the limit must be $0$ because the expression is $0$ in the whole interval except for $x=0$. This clearly implies that the limit must be $0$

Answer 3

For all $x>0$ we have that $x + x^2 > 0$. Secondly, you know that $\lim_{x\to 0^+}(x+x^2) = 0$, hence for all $0<\varepsilon < 1$ there exists an $x_\varepsilon>0$ such that $$0<x_\varepsilon+x_\varepsilon^2 <\varepsilon<1. $$

For any such $x_\varepsilon$ we have that $$\lfloor x_\varepsilon + x_\varepsilon^2\rfloor = 0.$$

Then we get that $$0\leq\frac{\lfloor x_\varepsilon + x_\varepsilon^2\rfloor}{x_\varepsilon} = \frac{0}{x_\varepsilon} = 0.$$