# Suppose $\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$,is $\lim f-g=0$?

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.

In calculus, there are three ways to approach a $\lim_\limits{x\rightarrow\infty}$ problem:

1. If the degree of the numerator > the degree of the denominator, such as in $\frac{x^3}{x^2}$, the limit always fails to exist (because it approaches either positive or negative infinity).

2. If the degree of the numerator < the degree of the denominator, such as in $\frac{x^2}{x^3}$, the limit is always 0.

3. If the degree of the numerator = the degree of the denominator, such as in $\frac{3x^2}{x^2}$, the limit is always the quotient of the coefficients of the highest degrees in the numerator and denominator. In the case of $\frac{3x^2}{x^2}$, $\lim_\limits{x\rightarrow\infty}$ $\frac{3x^2}{x^2}$ = $\frac 31$ = 3.

Thus, a rational function can obtain a limit of 1 as x approaches infinity if it satisfies the 3rd condition.

Since we are looking for a nonzero answer to $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)})$ where $\lim_\limits{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$, then a rational function with numerator $f(x)$ and denominator $g(x)$ must satisfy these conditions:

1. The quotient of the leading coefficients of the numerator and denominator must equal 1.
2. ${f(x)}$ $\neq$ ${g(x)}$
• Why? If ${f(x)}$ were equal to ${g(x)}$, then the $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)}) = 0$.

From here, we can name a few pairs of functions that would work.

• G's answer: $f(x) = x + 2$, $g(x) = x$, where $\lim_\limits{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$ and $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)}) = 2$

• $f(x) = x^2$, $g(x) = x^2 - 9$, where $\lim_\limits{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$ and $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)}) = 9$

• $f(x) = 5x^4 + 25$, $g(x) = 5x^4 - 75$, where $\lim_\limits{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$ and $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)}) = 100$

In each one, we notice that it is the constants that allow $\lim_\limits{x\rightarrow\infty} ({f(x)}-{g(x)})\neq 0$.

Consider $f(x)=x+2$ and $g(x)=x$. Or $f(x)=x+\sin x\cdot \ln x$ and $g(x)=x$

• Thanks, of course! so simple but I didnt thing about it – TheGeometer Oct 9 '16 at 19:33

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 $g(x)$.