Confusion Over Ratio Limits and Difference Limits

Question

I have two functions, $f(x)$: $$f(x)=\left(H\left(x\right)+\ln\left(H\left(x\right)^{\left(e^{H\left(x\right)}\right)}\right)\right)$$ and $g(x)$: $$g(x)=e^\gamma x\ln\left(\gamma+\ln x\right)+\frac{x}{\ln\left(\ln\left(x\right)\right)}$$ where $H(x)$ is the harmonic series.

What I find confusing is that $$\lim_{x \to \infty}\frac{f(x)}{g(x)}=1$$ but $$\lim_{x \to \infty}f(x)-g(x)\neq 1$$ Is there any way to consolidate these two facts?

Answer 1

Generally speaking, ratio limits are less sensitive than difference limits. As a rule, when you take a ratio limit of two functions $f\left(x\right)$ and $g\left(x\right)$ as $x$ tends to $c$ (where $c$ is a real number, or positive or negative $\infty$)—possibly from a particular direction (above/below)—the value of the limit reflects the balance between the dominant terms of $f\left(x\right)$ and $g\left(x\right)$.

Consider, for instance, the functions: $$f\left(x\right)=\frac{a}{\sqrt{x}}+\frac{b}{x}-c\ln x$$ $$g\left(x\right)=\frac{b}{x}$$ where $a,b,c$ are positive real numbers. Both $f\left(x\right)$ and $g\left(x\right)$ tend to positive $\infty$ as $x$ decreases to $0$. However: $$\lim_{x\downarrow0}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x\downarrow0}\left(\frac{a}{b}\sqrt{x}+1-\frac{c}{b}x\ln x\right)=1$$ but:$$\lim_{x\downarrow0}\left(f\left(x\right)-g\left(x\right)\right)=\lim_{x\downarrow0}\left(\frac{a}{\sqrt{x}}-c\ln x\right)=+\infty$$ The reason these two limits disagree has to do with the fact that, as $x$ decreases to $0$, the dominant terms of $f\left(x\right)$ and $g\left(x\right)$ are both $\frac{b}{x}$, because $\frac{b}{x}$ grows faster as $x$ decreases to $0$ than either $\frac{a}{\sqrt{x}}$ or $-c\ln x$. On the other hand, because $f\left(x\right)$ contains the terms $\frac{a}{\sqrt{x}}$ and $-c\ln x$—both of which grow as $x$ tends to $0$—which are not cancelled out by terms in $g\left(x\right)$, the difference limit explodes to $\infty$.

One trick you can sometimes use is the following: if you have functions $f\left(x\right)$ and $g_{0}\left(x\right)$ so that:$$\lim_{x\rightarrow c}\frac{f\left(x\right)}{g_{0}\left(x\right)}=1$$ then this tells you that the dominant terms of $f\left(x\right)$ and $g_{0}\left(x\right)$ as $x\rightarrow c$ are identical. If you can find a function $g_{1}\left(x\right)$ so that: $$\lim_{x\rightarrow c}\frac{f\left(x\right)-g_{0}\left(x\right)}{g_{1}\left(x\right)}=1$$ then that means that the term of $f\left(x\right)$ of dominant growth after $g_{0}\left(x\right)$ is $g_{1}\left(x\right)$. If you can keep finding $g_{2}\left(x\right)$: $$\lim_{x\rightarrow c}\frac{f\left(x\right)-g_{0}\left(x\right)-g_{1}\left(x\right)}{g_{2}\left(x\right)}=1$$ and $g_{3}\left(x\right)$ and $g_{4}\left(x\right)$ and so on, you will eventually construct an asymptotic expansion for $f\left(x\right)$:$$f\left(x\right)\sim g_{0}\left(x\right)+g_{1}\left(x\right)+g_{2}\left(x\right)+\cdots\textrm{ as }x\rightarrow c$$ This is basically the same as taking a (generalized) taylor series expansion of $f\left(x\right)$ at $x=c$. Once you get the complete expansion (either terminating after finitely many $g_{n}\left(x\right)$s, or an infinite series of $g_{n}\left(x\right)$s), the resultant function will then cancel out $f\left(x\right)$ in a difference limit as $x\rightarrow c$.

On the other hand, if you have two functions $f\left(x\right)$ and $g\left(x\right)$ which blow up at $x=c$, but their difference is finite-valued at $x=c$, then that means that the singular part (the divergent part) of $f\left(x\right)$ at $x=c$ is exactly the same as the singular part of $g\left(x\right)$. A very important example of this is with the Riemann Zeta Function:$$\zeta\left(x\right)=\sum_{n=1}^{\infty}\frac{1}{n^{x}}$$ and the function $\frac{1}{x-1}$. Even though both of these functions blow up as $x\rightarrow1$, their difference is continuous at $x=1$, where it equals a famous finite quantity, $\gamma$, the Euler-Mascheroni constant:$$\lim_{x\rightarrow1}\left(\zeta\left(x\right)-\frac{1}{x-1}\right)=\gamma$$ Thus, this is an example where both the ratio and difference limits are finite:$$\lim_{x\rightarrow1}\frac{\zeta\left(x\right)}{\frac{1}{x-1}}=\lim_{x\rightarrow1}\left(x-1\right)\zeta\left(x\right)=1$$ meaning that not only do $\zeta\left(x\right)$ and $\frac{1}{x-1}$ have the same dominant terms as $x\rightarrow1$, but also—since the difference limit is finite—that there are no other singular terms of $\zeta\left(x\right)$ at $x=1$ except for $\frac{1}{x-1}$. Indeed, $\zeta\left(x\right)-\frac{1}{x-1}$ is analytic (represented by a taylor series) about $x=1$.