TITLE: Suppose $\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$,is $\lim f-g=0$?
QUESTION [5 upvotes]: I was wondering if it is possible that $\lim_\limits{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$, but $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)})\neq 0$ or even be non existent. It seems intuitive that the result will always be zero and indeed is easy to prove when both the limits of $f$ and $g$ exist, however I can't prove it in the case the limits do not exist. So I am not sure if it possible that the result is not zero or the limit to be nonexistent.  
Thanks in advance.

REPLY [3 votes]: If $f(x)/g(x) = 1$ then $f(x) = g(x)$ and $f(x) - g(x) = 0$ and many many students so desperately want this logical inference to be valid in limit operations also. Such a desire stems from the resistance to ditch the algebraical approach of $+,-,\times,/,=$ when one starts studying calculus.
When we deal with limits it is essential to understand that they are a non-algebraic concept based on order relations of $<, >$ and one should expect that they behave differently. Thus one should not be surprised if $$\lim_{x \to \infty}\frac{f(x)}{g(x)} = 1$$ does not imply that $$\lim_{x \to \infty}\{f(x) - g(x)\} = 0$$ We can see that $$\lim_{x \to \infty}\{f(x) - g(x)\} = \lim_{x \to \infty}g(x)\left(\frac{f(x)}{g(x)} - 1\right) = \lim_{x \to \infty}g(x) h(x)$$ where $h(x) = f(x)/g(x) - 1$. Now the limit of $h(x)$ is $0$ but it does not necessarily mean that limit of $g(x)h(x)$ is also $0$. One can't make an inference like this without knowing further about $g(x)$.