What is the value of this limit? Does the value not exist or is this limit just indeterminate? Does $\;\lim\limits_{x\to\infty}\dfrac2{5x|\sin x|}\;$ equal DNE (does not exist) or converge to $\,0$ ?
Although the $5x$ in the denominator approaches $\infty$, shouldn't the limit be DNE since at an arbitrary value of $x$, $|\sin x|$ will be $0.0000000000...1$ making the denominator not $\infty$ and the limit be something else other than $0$. Since the limit is not always $0$, must it not exist?
I had a conversation with my math teacher and he said it's definitely indeterminate as there may be a value that makes it not equal to $0$ but I have to prove that there are an infinite amount numbers that will make this not equal to $0$. If you can prove this, try proving it for natural numbers.
 A: We will prove that there exists an increasing and unbounded above sequence $(x_n)$ of positive real numbers such that $\;f(x_n)\neq0\;$ where $$f(x)=\dfrac2{5x|\sin x|}\,.$$
Proof:
For any $\;n\in\Bbb N\,,\;$ let $\;g_n\!:\!\left[2\pi n,\dfrac{\pi}2\!+\!2\pi n\right]\to\Bbb R\;$ be the real function defined as follows :
$g_n(x)=x|\sin x|\quad$ for all $\;x\in\left[2\pi n,\dfrac{\pi}2\!+\!2\pi n\right].$
Since $\;g_n\;$ is a continuous function on its domain, $\;g_n(2\pi n)=0\;$ and $\;g_n\!\left(\dfrac{\pi}2\!+\!2\pi n\right)=\dfrac{\pi}2\!+\!2\pi n>1\,,\,$ by using the Intermediate value theorem , we get that there exists $\;x_n\in\left]2\pi n,\dfrac{\pi}2\!+\!2\pi n\right[\;$ such that $\;g_n(x_n)=x_n|\sin x_n|=1\,.$
In this way we get a sequence $(x_n)$ of positive real numbers such that
$f(x_n)=\dfrac2{5x_n|\sin x_n|}=\dfrac25\neq0\,.$
Moreover, the sequence $(x_n)$ is increasing and unbounded above, indeed
$\color{blue}{2\pi n<x_n}<\dfrac{\pi}2\!+\!2\pi n<2\pi(n+1)\color{blue}{<x_{n+1}}\quad$ for any $\;n\in\Bbb N\,.$
A: You are right in your effort or "mathematical guess": the limit does not exist because the function isn't defined in infinite points when $\;x\to\infty\;$, namely: all the points of the form $\;x=k\pi\;,\;\;k\in\Bbb N\;$ . This means that in any "neighborhood of infinity" (which formally means: for any $\;x>R\;,\;\;0<R\in\Bbb R$), there exist infinite points in that neighborhoodf for which the function does not even exist.
