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In the interval \((-2 \pi, 0)\), the function \(f(x)=\sin \left(\frac{1}{x^3}\right)\)
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The correct answer is:
changes sign infinitely many times
Hint : \(x \in(-2 \pi, 0), f(x)=\sin \left(\frac{1}{x^3}\right)\)
\(\begin{aligned}
& \because-2 \pi < x < 0 \\
& \Rightarrow-8 \pi^3 < x^3 < 0 \\
& \Rightarrow-\infty < \frac{1}{x^3} < -\frac{1}{8 \pi^3}
\end{aligned}\)
Hence, \(\sin \left(\frac{1}{x^3}\right)\) will take all values from -1 to 1 , and will change its sign infinitely many times since it is a periodic function
\(\begin{aligned}
& \because-2 \pi < x < 0 \\
& \Rightarrow-8 \pi^3 < x^3 < 0 \\
& \Rightarrow-\infty < \frac{1}{x^3} < -\frac{1}{8 \pi^3}
\end{aligned}\)
Hence, \(\sin \left(\frac{1}{x^3}\right)\) will take all values from -1 to 1 , and will change its sign infinitely many times since it is a periodic function
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