Find the limit of the sequence $x_n = \sqrt[n]{2^n - \sin(n)}$ and explain your answer. The most common way to solve this I think is using sandwich rule, but I cannot find the lower value.Or if there is a better way to do this?
Showing the limit of $x_n = \sqrt[n]{2^n - \sin(n)}$
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sequences-and-series
analysis
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0is your sequence $$a_n=n\sqrt{2^n-\sin(n)}$$? – 2017-02-04
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0Do I have to explain the answer? – 2017-02-04
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2Stating just a question is somewhat considered a rude manner, as it seems like you're ordering people here to solve your problem. Anyway it would be better to give us your thoughts on this problem and tell us where you encounter a problem. – 2017-02-04
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0the limit doesn' exist – 2017-02-04
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0There are many questions about mathematical analysis ... The title is so bad :| – 2017-02-04
2 Answers
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$$x_n = \sqrt[n]{2^n - \sin(n)}\\ x_n = \sqrt[n]{2^n (1- \dfrac{\sin(n)}{2^n})}=\\ x_n = 2\times\sqrt[n]{ (1- \dfrac{\sin(n)}{2^n})}=\\ \lim_{n \rightarrow \infty}x_n = \lim_{n \rightarrow \infty}2\times\sqrt[n]{ (1- \dfrac{\sin(n)}{2^n})}=2\\\text{because} \\ \lim_{n \rightarrow \infty} \dfrac{\sin(n)}{2^n}\to 0$$
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Check to see that $$2^n-1 \leq 2^n-\sin(n) \leq 2^n+1$$
and that $1 From here, my hint is to use the "squeeze theorem" and evaluate
$$\lim_{n \to \infty} \left( 2^n \pm 1 \right)^{1/n}$$ which is much easier.