In the interval (-2 pi, 0), the function f(x) | vedclass

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(D) Given the interval $x \in (-2 \pi, 0)$.As $x$ approaches $0$ from the left, $x^3$ approaches $0$ from the negative side.Therefore, $\frac{1}{x^3}$

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changes sign infinitely many times

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(D) Given the interval $x \in (-2 \pi, 0)$.As $x$ approaches $0$ from the left, $x^3$ approaches $0$ from the negative side.Therefore, $\frac{1}{x^3}$ approaches $-\infty$.Since the function $f(x) = \sin \left(\frac{1}{x^3}\right)$ oscillates between $-1$ and $1$ as its argument $\frac{1}{x^3}$ approaches $-\infty$, the function will cross the $x$-axis infinitely many times in any neighborhood of $0$.Thus, the function changes sign infinitely many times in the interval $(-2 \pi, 0)$.

In the interval $(-2 \pi, 0)$, the function $f(x) = \sin \left(\frac{1}{x^3}\right)$

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