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

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.

• 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 $m,n$. – Clayton Apr 12 '19 at 3:41
• Just out of curiosity, what is the larger problem? – Ovi Apr 12 '19 at 3:43
• @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). – user663014 Apr 12 '19 at 3:48
• This question may help. – Clayton Apr 12 '19 at 3:51
• 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) – user663014 Apr 12 '19 at 3:55

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.