I am working with the supposition that $\lim\limits_{x\to \infty}{f(x)}=\infty$. I have rewritten it as $$\forall R>0\;\exists r>0\;\text{ such that }\;|x|>r\Rightarrow |f(x)|>R.$$ I want to turn this "$p\Rightarrow q$"-type supposition into a $\neg q\Rightarrow \neg p$ one, but I am not sure how to negate it. Is $$\forall R>0\;\exists r>0\;\text{ such that }\;|f(x)|\leq R\Rightarrow |x|\leq r$$ correct?

  • $\begingroup$ The function is $f: \mathbb{R}^m \to \mathbb{R}^n$ $\endgroup$
    – khalelbm
    Sep 21 at 12:34

1 Answer 1


You have hidden quantifiers that aren't being accounted for. Let us take a closer look at "$\lim_{x\to\infty}f(x)\neq\infty$".

$$(\forall R>0)(\exists r>0)(|x|>r\implies|f(x)|>R)$$

While this seems sensible enough, the variable $x$ here is free. This is a question about a single value of $x$. If I were to plug in $x=0$ then we would have that this always holds (for every function $f$). Instead, what is really happening is

$$(\forall R>0)(\exists r>0)(\forall x)(|x|>r\implies|f(x)|>R)$$

Now you can use modus tollens to obtain the form that you seek.

$$(\forall R>0)(\exists r>0)(\forall x)(|f(x)|\leq R\implies |x|\leq r)$$

I think that you may want to do away with your absolute values, though. $\lim_{x\to\infty}f(x)=\infty$ is that in the positive $x$ direction $f$ eventually becomes arbitrarily large. Therefore, it would look more like this:

$$(\forall R>0)(\exists r>0)(\forall x)(f(x)\leq R\implies x\leq r)$$


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