Let $f(x)$ and $g(x)$ be two functions such that $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$$ Does it imply:

  1. $\lim_{x\to\infty}\frac{g(x)}{f(x)}=1$
  2. $\lim_{x\to\infty}\left(f(x)-g(x)\right)=0$

To me it seems like both are true, but the second is obviously false: $$\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)=1/2$$ but why? If both functions are basically the same far enough in the number line why does the limit not approach $0$. I don't know how to give a rigorous answer (rather than examples)

Are the answers different in these two cases?

  1. $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\infty$
  2. $f(x)$ and $g(x)$ are bounded


  • 1
    $\begingroup$ The first is true simply because the function $x \mapsto 1/x$ is continuous at $x=1$. $\endgroup$ – Michał Miśkiewicz Apr 30 '17 at 11:30
  • 7
    $\begingroup$ For the second one: If I have 1 billion dollars and you have 1 billion dollar plus 1 dollar, we are essentially equally rich (i.e. the ratio of your wealth to mine is essentially 1). But the difference between us is still 1 dollar. Is that dollar significant? In percentage terms, it isn't (i.e. ratio is close to 1). But in monetary terms is still 1 dollar $\endgroup$ – Ant Apr 30 '17 at 12:28
  • $\begingroup$ @Ant wonderful explanation I get it now. Thanks $\endgroup$ – augustoperez Apr 30 '17 at 13:33

Here's a hint: Note that $$ f(x)-g(x) = \biggl(\frac{f(x)}{g(x)}-1\biggr)g(x) .$$ Then look at the two factors on the right hand side. The first approaches zero, by assumption. What might happen if the other factor is bounded? Not bounded?

Another, less rigorous but perhaps more entertaining way to look at it: It's the difference between relative differences and absolute differences. Take any two of the five richest people on the planet. In relative terms, they are about equally rich. But if I could have the difference between their absolute wealths, I could certainly retire tomorrow. What might be a small difference to them is a huge difference to me.



$$\lim_{+\infty}\frac {f (x)}{g (x)}=1$$

means that for $x $ great enough,

$$f (x)=g (x)(1+\epsilon (x)) $$ with $\epsilon (x)\to 0$ when $x\to +\infty$. thus

$$f (x)-g (x)=g (x)\epsilon (x) $$ from here, we see that $$\lim_{+\infty}(f (x)-g (x)) $$ depends strongly on $\epsilon (x) $.


Note that $f(x)-g(x)=f(x)\left(1-\frac{g(x)}{f(x)}\right)$. Although $\lim_{x\to\infty}\left(1-\frac{f(x)}{g(x)}\right)=0$, but if $\lim_{x\to\infty}f(x)=\infty$, we cannot tell whether $\left[f(x)\left(1-\frac{g(x)}{f(x)}\right)\right]$ tends to $0$ or not.


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