# Confusion Over Ratio Limits and Difference Limits

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?

• Have you tried taking lots of $\log$'s? Note that $\log x$ is a strictly increasing function. Commented Jun 18, 2019 at 3:52
• I'm sorry if I don't understand but how does that help me? Commented Jun 19, 2019 at 15:51
• Monotonicity of $\log g(n)$ implies monotonicity of $g(n)$. So does $\log \log g(n)$, etc. I'm not sure if it will immediately answer your question, but it is the first obvious thing to try. Commented Jun 19, 2019 at 16:24

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 $$x=1$$.