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Does $\lim_{x\to 0} f(x) \geq \lim_{x\to 0} c \cdot g(x)$ imply $f(x) \geq c \cdot g(x)$?

Here, $c$ is a positive constant. Essentially, I want to know whether or not I can drop the limit. This is part of a larger proof I'm working on, and if I'm allowed to drop the limit, I'll be done with the proof.

I think it's allowed, simply because I wouldn't know how to solve the larger problem that I'm working on if it weren't allowed. I just want to make sure / understand when this is a valid thing to do. Thanks so much.

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  • $\begingroup$ Certainly not; consider $f(x)=-|x|$ and $g(x)=|x|$. We have equality in the limit and a strict reverse inequality for all other $x$, regardless of the positive constant $c$. If you want to impose that $f$ is positive, consider $f(x)=x^n$ and $g(x)=x^m$ for suitable $m,n$. $\endgroup$ – Clayton Apr 12 '19 at 3:41
  • $\begingroup$ Just out of curiosity, what is the larger problem? $\endgroup$ – Ovi Apr 12 '19 at 3:43
  • $\begingroup$ @Ovi Let $U$ be an open subset of $\mathbb{R}^{n}$ and suppose that the continuously differentiable mapping $F : U \rightarrow \mathbb{R}^{n}$ is stable. Prove that for each $x \in U$, the derivative matrix $DF(x)$ is invertible. (Hint: Use the first-order approximation theorem). $\endgroup$ – user663014 Apr 12 '19 at 3:48
  • $\begingroup$ This question may help. $\endgroup$ – Clayton Apr 12 '19 at 3:51
  • $\begingroup$ Yeah I saw that post, and I got that far as well. Also, I know that an $n \times n$ matrix $A$ is invertible if and only if there is $c > 0$ such that $||Ah|| \geq c||h||$. But in order to use that result, we'd need to drop the limit as $h\to 0$ from each side (which is why I asked this question) $\endgroup$ – user663014 Apr 12 '19 at 3:55
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Take $f(x)=x$ and $g(x)=x^2$.

We see $$\lim_{x\rightarrow0}f(x)\geq\lim_{x\rightarrow0}g(x),$$ but $x\geq x^2$ is wrong.

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