I have to evaluate the following limit $$\lim_{n\to\infty}\frac{n^3\sin(n!)}{n^5+1}$$ I know the answer is zero and since $$\lim_{n\to\infty}\frac{1}{n^5+1}=0$$ I want to show that $n^3\sin(n!)$ is bounded. I know for every $n\in\mathbb{N}$, $$-1\leq\sin(n!)\leq1$$ But I don't know what can I do with $n^3$. When I tried to use WolframAlpha, it says $$-69\leq n^3\sin(n!)\leq59$$ However, I have no idea how to get these values.
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3I’d try another method since $n^3 \sin(n!)$ is very likely not bounded. Can you compare the absolute value of the numerator and the denominator in your fraction? – Adam Z Sep 18 '23 at 01:42
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5Who knows how $n^3\sin(n!)$ is behaved. All you need is $-1\le\sin(n!)\le1$ though, since $\frac{n^3}{n^5+1}\to0$. – coiso Sep 18 '23 at 01:45
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2$\left|\frac{n^3\sin(n!)}{n^5+1}\right| \leq \frac{n^3}{n^5+1} < \frac{n^3}{n^5} = \frac{1}{n^2} $ – Will Jagy Sep 18 '23 at 02:28
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1I'm not sure why this was down-voted. – Theo Bendit Sep 18 '23 at 03:37
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1@TheoBendit No quality control on down-voting. – user2661923 Sep 18 '23 at 03:50
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1See https://math.stackexchange.com/q/677719/27978. I suspect trying to show that $n^3 \sin (n!)$ is bounded would be difficult. See the related MO commentary. – copper.hat Sep 18 '23 at 04:01
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I doubt Wolfram Alpha says $-69\leq n^3\sin(n!)\leq 59$ – jjagmath Sep 18 '23 at 12:01
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@MarianoSuárez-Álvarez I agree that W|A must be used with caution. In this case however, I can't imagine what the OP input on W|A to get that result, because it certainly doesn't give that bounds. – jjagmath Sep 18 '23 at 12:05
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@jjagmath you can doubt about it, but yes, Wolfram Alpha says it. – mvfs314 Sep 18 '23 at 14:20
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@MarianoSuárez-Álvarez fortunately I know what I'm doing and that's why I came here to see what's the problem. – mvfs314 Sep 18 '23 at 14:22
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@mvfs314 I shared the link where you can see it's not the case. Why don't you share the link of your search in Wolfram Alpha which says that? – jjagmath Sep 18 '23 at 18:21
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I'm not sure if your question is show that $n^3\sin(n!)$ is bounded or you are trying to find the limit of $\frac{n^3\sin(n!)}{n^5+1}$ when $n{\to}$$\infty$ but,
if you are trying to find the limit of the above function then it's not the best way to show that $n^3 \sin(n!)$ is bounded instead you can show that $ \sin(n!)$ is bounded by one,
i.e $|\sin(n!)|\leq 1 $and $\lim_{n \to +\infty} \frac{n^3}{n^5+1}=0$ so $\lim_{n \to +\infty}\frac{n^3\sin(n!)}{n^5+1}=0$ .
Ricky
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Mahmoud Mrowi
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If you precede trigonometric functions' names with a backslash, MathJax/LaTeX will know they are symbols. Then it will render them in an upright font and with spacing appropriate for functions' names, instead of an amorphous blob of italic, variable-like, letters. Compare
a sin x→ $a sin x$ vs.a\sin x→ $a\sin x$. – CiaPan Sep 18 '23 at 12:08