How can I evaluate the limit?
$$\lim_{n \rightarrow \infty} \sin (n {\pi} {\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}})$$
I think the limit is $0$ but I fail to work it out properly. Please help me.
Thank you in advance.
Compute $\lim_{n \rightarrow \infty} \sin (n {\pi} {\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}})$
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calculus
limits
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0Is $\pie$ supposed to be $\pi$ or $\pi\cdot e$? – 2017-01-06
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0$\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}\to n+1$ as $n\to\infty$. Note $f(x)=\sin(x)$ is a continues function, thus $$\lim_{n\to\infty}\sin(...)=\sin (\lim_{n\to\infty} ...) $$ – 2017-01-06
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0How can I show this? – 2017-01-06
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2$$\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}=\sqrt[3] {(n+1)^3+n-6}=(n+1)\sqrt[3]{1+\frac{n-6}{(n+1)^3}}$$ – 2017-01-06
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0i.e. according to your comment the limit becomes $$\lim _{n \rightarrow \infty} \sin (n(n + 1) {\pi})$$ which is simply $0$. – 2017-01-06
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0Is this over $\mathbb{N}$? – 2017-01-06
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0I see this question is marked as a duplicate, I do not agree with that. The question is similar, and so is the resolution method. But the answer is different. Moreover, the arguments used in answers of $\sqrt[3]{n^3+1}$ do not detail why this $\varepsilon_n\to 0$. And in this precise case, the Taylor expansion shows that a constant term in $\frac{1}{3}$ appears, it is not just $0$ as in the other problem ! The comments I see about continuity of sinus seem to obliviate that the real issue is about what's inside the sinus and the absolute need of Taylor expanding the cubic root. – 2017-01-07
2 Answers
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Hint
$\sin(n^2\pi\sqrt[3]{1+\frac{3}{n}+\frac{4}{n^2}-\frac{5}{n^3}})$ and use Taylor expansion of the cubic root.
Solution : $\frac{\sqrt 3}{2}$ because of parity ;-)
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0What do you mean with DL ? – 2017-01-06
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0Typo, I corrected. DL=sorry french acronym for Taylor developpement. – 2017-01-06
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0Yes, your answers sound very French – 2017-01-06
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To be as exact as possible, do not use that some limit is $n+1$, but use the difference to that sequence:
$$ \sin(n\pi(n+1+ε_n))=(-1)^{n(n+1)}\sin(n\pi\varepsilon_n) $$ with \begin{align} ε_n&=\sqrt[3]{n^3+3n^2+4n-5}-(n+1) \\ &=(n+1)\left(\sqrt[3]{1+\frac{n-6}{(n+1)^3}}-1\right) \\ &=\frac1{3(n+1)}+O\left(\frac1{(n+1)^2}\right) \end{align} As $n(n+1)$ is always even and $nε_n\to\frac13$, the sequence converges towards $\sin(\frac\pi3)$
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0According to my calculation ${n \epsilon_{n}} \rightarrow \frac {-5} {3}$ as $n \rightarrow \infty$.But the limit is same which is $\frac {\sqrt 3} {2}$. – 2017-01-06
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0I do not see how that is possible based on the above computation. As the difference is -2 or $-2\pi$ in the sine argument, it does indeed make no difference. $-$ But note that there was an error in the power of $(n+1)$ in the second cube root. The last line remains the same as $(n+1)·\frac13·\frac{n-6}{(n+1)^3}=\frac1{3(n+1)}-\frac{7}{3(n+1)^2}$. – 2017-01-06