I watched a youtube video explaining how to get
$\lim \limits_{x \to \infty} \left( \frac{\sin{(2x)}}{x} \right)$
He applied the properties of limits such that
$\lim \limits_{x \to \infty} \left( \frac{1}{x} \right) \cdot \lim \limits_{x \to \infty} \sin{(2x)}$
In which, after evaluation,
$\lim \limits_{x \to \infty} \left( \frac{1}{x} \right) = 0$
and
$\lim \limits_{x \to \infty} \sin{(2x)}$ = DNE.
So, it went
$ 0 \cdot DNE$
Is this always the case, especially for transcendental functions like trigonometric functions?
If $ 0 \cdot DNE$ can be equal to zero, then why is $ 0 \cdot \infty$ indeterminate?
I understand that infinity is not a number but a concept and there's no way to determine $ 0 \cdot \infty$ to be anything else. But why is DNE a special case if it is not a number too?
Thanks!
This is wrong. The assertion$$\lim_{x\to\infty}\bigl(f(x)g(x)\bigr)=\left(\lim_{x\to\infty}f(x)\right)\left(\lim_{x\to\infty}g(x)\right)\tag1$$holds when both limits of the RHS of $(1)$ exist, but that's not the case here. However, it is true that if $\lim_{x\to\infty}f(x)=0$ and if $g$ is a bounded function, then $\lim_{x\to\infty}\bigl(f(x)g(x)\bigr)=0$, and this can be used here.