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$$n^{3}\sin\left(\frac{5}{n^{3}}\right)$$ I need to figure the limit if it converges or state if it goes to $$ \pm\infty $$ or simply "divergent" if it diverges but not to infinity.

Looking at the sequence I can tell that as n goes to infinity $$n^{3} \to \infty $$ so it is divergent and $$\sin\left(\frac{5}{n^{3}}\right)$$ is convergent from using the p-series comparison test with $$\frac{1}{n^{3}}$$ so a divergent series multiplying a convergent series would overall make a divergent series ? Or because $$\frac{5}{n^{3}} $$ is approaching $0$ as $n$ increases and the $\sin(0) = 0$ would the whole series converges to $0$?

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$$\lim_{n\to\infty}n^3\sin\frac{5}{n^3}=5\lim_{n\to\infty}\frac{\sin\frac{5}{n^3}}{\frac{5}{n^3}}=5\lim_{x\to0}\frac{\sin x}{x}=5$$

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What you should use this the fact that

$$\sin(x)\sim_{x\to 0} x.$$

So

$$\sin(5/n^3)\sim_{n\to \infty} 5/n^3$$

since $5/n^3\to 0$ when $n\to \infty$.

So $$n^3\sin(5/n^3)\sim_{n\to\infty}n^3\frac 5{n^3}=5.$$

Finally you can conclude that this sequence converges to $5$ for $n\to \pm\infty$.

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write your term in the form $$5\frac{\sin\left(\frac{5}{n^3}\right)}{\frac{5}{n^3}}$$

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