In the interval$ (-2\pi, 0)$, the function$ f(x) = \sin (1/x^3)$ changes its sign how many times?

Question

options given are : (A) Never changes sign (B) Changes sign only once (C) Changes sign more than once, but a finite number of times (D) Changes sign $\infty$ number of times

I have used google to draw the plot of $ f(x) = \sin (1/x^3)$ and my conclusion based on my intuition is (D)=$\infty$ no. of times.

But proper mathematical reasoning is what I require to support the answer.

Answer 1

The function is $1$ at all points such that $\frac{1}{x^3}=2k\pi+\frac{\pi}{2}$ for some $k \in \mathbb{Z}$ $\Leftrightarrow x=\frac{1}{(2k\pi+\frac{\pi}{2})^\frac{1}{3}}=g(k)$. Notice that $g(k)$ is in the interval $(-2\pi,0)$ for all sufficiently small $k$ (small as in negative but having large modulus) which shows that your initial function hits $1$ infinitely many times in your interval. You can similarly show that it is infinitely many times equal to $-1$. Now argue by continuity that it changes sign infinitely often.

Answer 2

$\sin x$ changes of sign for $x \in \{k\pi ; k\in \mathbb Z\}$. Hence $\sin\left(1/x^3 \right)$ changes of sign for $x \in A=\{ \frac{\pm 1}{\sqrt[3]{k}}; k \in \mathbb N \}$. And the cardinality of $A \cap (-2\pi, 0)$ is infinite.