# Function that makes limit finite

Question: Let $$n > 0$$. How can I find a function $$f:\mathbb{N}\rightarrow\mathbb{R}^+$$ such that $$\lim_{n\to\infty} \frac{f(n)^2}{n} \log \left(\frac{f(n)}{n}\right) = L$$ with $$0?

Background: The term above appears in my research on subexponential bounds for binary words containing a limited number of ones. I have been able to elimate all other terms, but I am stuck with this one.

What I tried so far: I applied L'Hôpital's rule to get $$\lim_{n\to\infty} \frac{\log\left(\frac{f(n)}{n}\right)}{\frac{-n}{f(n)^2}} = \lim_{n\to\infty} \frac{\frac{f'(n)}{f(n)}-\frac{1}{n}}{\frac{1}{f(n)^2}-\frac{2nf'(n)}{f(n)^3}}$$

which got rid of the $$\log()$$. Since the limit should be finite, it seems to me that $$\lim_{n\to\infty} \frac{f(n)}{\sqrt{n}} < \infty$$, but I haven't been able to come up with an $$f(n)$$ that doesn't lead to $$L=0$$.

• By $f(n)^2$ do you mean $(f(n))^2$ or $f(n^2)$ ? Apr 6, 2022 at 13:48
• Also, does $f$ have to be a function with integer outputs? Or it can have real, non-integer outputs? Lots to clarify in the question in my opinion... Apr 6, 2022 at 14:01
• @AdamRubinson with $f(n)^2$, I mean $f(n)f(n)=(f(n))^2$ Apr 6, 2022 at 15:02

Take $$f(n) = n+1$$. Then \begin{align*} \lim\limits_{n\to\infty} \frac{f(n)^2}{n}\log\bigg(\frac{f(n)}{n}\bigg) = \lim\limits_{n\to\infty} \bigg(n + 2 + \frac 1n\bigg)\log\bigg(1 + \frac 1n\bigg) = \lim\limits_{n\to\infty} \frac{n + 2 + \frac 1n}{n}\cdot n\log\bigg(1 + \frac 1n\bigg) = 1. \end{align*} You can get any other number $$L>0$$ by taking $$f(n) = n+L$$.
I'll assume that $$f$$ is a function which can output real values.
For each $$n\in\mathbb{N},$$ define $$g_n:[n,\infty)\to\mathbb{R};\ g_n(x) = \frac{x^2}{n} \log\left(\frac{x}{n}\right).$$
One can readily check that $$g_n(x)$$ is a continuous, increasing function in $$x$$, with range $$[0,\infty).$$
Therefore (by IVT), for each $$n,\ \exists\ x_n\in [n,\infty)\$$ such that $$g_n(x_n)= \frac{{x_n}^2}{n} \log\left(\frac{x_n}{n}\right) = L.$$
Let $$f(n) = x_n$$ for each $$n\in\mathbb{N},$$ and we have completed our construction of $$f.$$