Strange behavior of the function $f(x)=\frac{x^2+\ln(\cos(x))}{x^4}$ when $x\to +\infty (-\infty)$

Question

I have explained the partial graph of this function

$$f(x)=\frac{x^{2}+\ln(\cos(x))}{x^{4}}$$

for my students of an high school. Considerated that the domain is given by $\cos(x)>0 \ \wedge x \neq 0$, I obtain

$$\text{dom} \ f(x)=\left]-\frac{\pi}{2},0\right[\cup \left]0,\frac{\pi}{2}\right[$$ without periodicity. The calculus of the limits are easy. My question is:

1. $$f(x)=\frac{x^{2}+\ln(\cos(x))}{x^{4}}=\frac1{x^2}+\frac{\ln(\cos(x))}{x^4}$$

I have thought the solve

$$\lim_{x\to +\infty}\frac{\ln(\cos(x))}{x^4}$$

using the squeeze theorem

$$-\frac{1}{x^4}\leq \frac{\ln(\cos(x))}{x^4} \leq \frac{1}{x^4}$$ to have

$$\lim_{x\to +\infty}\frac{\ln(\cos(x))}{x^4}=0$$

Are there other solutions to solve this limit?

At the end using Desmos tool I have a strange behavior when $x\to \infty$.

What exactly happens in the area shown in the illustration?

Answer 1

Some people might say right away that this limit is undefined because the function is never defined in an interval of the form $[B,\infty)$. However, the limit makes sense under a more general definition which imposes the condition $x\in D$ where $D$ is the domain of the function $f$. Still, the limit doesn't exist.

Note that your inequalities are wrong. $-1\leq \cos(x) \leq 1$ doesn't imply $-1\leq \ln(\cos(x)) \leq 1$. In fact, $\ln(\cos(x))$ can be an arbitrarily small negative number.

So assume to the contrary that the limit exists. Since $\lim_{x\to\infty}\frac{x^2}{x^4}=0$, it follows that $\lim_{x\to\infty}\frac{\ln(\cos(x))}{x^4}$ exists. Let $g(x)=\frac{\ln(\cos(x))}{x^4}$ defined on $D$. Consider the sequence $a_n=\frac{\pi}{3}+2\pi n\to \infty$ as $n\to\infty$. $$g(a_n)=\frac{\ln(1/2)}{\left(\frac \pi 3+2\pi n\right)^4}\to 0\,\,(n\to \infty)$$ which means that $\lim_{x\to\infty}g(x)=0$. The definition of limit at infinity implies that there is a number $N>0$ such that when $x>N$ and $x\in D$, we have $|g(x)|<1 \Leftrightarrow -1<g(x)<1$ (choosing $\varepsilon=1$). Consider $a=\frac{\pi}{2}+2\pi N > N$. We have $$\lim_{x\to a^{-}}g(x)=-\infty $$ because $\ln(\cos(x))\to -\infty$ as $x\to a^{-}$ (note this is a one-sided limit with $x<a$). So $-1<g(x)<1$ is false in some interval $(a-\delta,a)$, and we have a contradiction.

This shows that $f(x)$ oscillates between $-\infty$ and $0$. However, the interval where the function drops to $-\infty$ is very small, basically imperceptible on the plot. There are, in fact, many such examples where a function behaves too "wildly" for graphing software / calculators to handle properly.