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$$\lim_{n\to\infty}\sqrt[n]{3n^3+(-1)^nn+4}$$ $$\le\lim_{n\to\infty}\sqrt[n]{4n^3+4}$$ $$=\lim_{n\to\infty}\sqrt[n]4\sqrt[n]{n^3+1}=1$$ I tried to use the squeeze theorem to find the limit of the first expression, however, I couldn't think of anything for the lower bound. Can someone give me an idea on how to progress. thank you

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  • $\begingroup$ i can not read your limit $\endgroup$ – Dr. Sonnhard Graubner May 12 '17 at 15:51
  • $\begingroup$ sorry, it says n to infinity $\endgroup$ – zee May 12 '17 at 15:53
  • $\begingroup$ Why do you think $\sqrt[n] {3n^3 + (-1)^n*x* n+4} \le \sqrt[n] {3n^3 + (-1)^n n+4}$ ? That's obviously not true for even $n$. $\endgroup$ – fleablood May 12 '17 at 15:56
  • $\begingroup$ @fleablood The "×" looked so much like $x$. My edit is my best guess as to what the actual limit was. $\endgroup$ – Parcly Taxel May 12 '17 at 15:57
  • $\begingroup$ yes @ParclyTaxel is right sorry for the confusion $\endgroup$ – zee May 12 '17 at 15:59
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I would say the lower bound is the easier one. We have $$3n^3+(-1)^nn+4>3n^3-n=n(3n^2-1)$$ which is the product of two increasing sequences, so we get a lower bound of $1\cdot(3-1)=2$. I would go slightly further than your upper bound to be safe, though: $$3n^3+(-1)^n+4\le4n^3+4\le4n^3+4n^3=8n^3.$$ Taking $n^\text{th}$ roots we have $$\sqrt[n]{2}\le\sqrt[n]{3n^3+(-1)^nn+4}\le\sqrt[n]{8}(\sqrt[n]{n})^3$$ and so the result follows by the squeeze theorem.

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Hint: Take $\sqrt[n]{3n^{3}+(-1)^{n}n+4}=1+c_{n}$. Then try to show that $c_{n}$ goes to $0$.

As the hint was not helpful i will try to provide a more elaborate asnwer. As per the hint we have $3n^{3}+(-1)^{n}n+4=(1+c_{n})^{n}\geq 1+\frac{1}{4!}n(n-1)(n-2)(n-3)c_{n}^{4}$ $\Rightarrow$ $c_{n}^{4}\leq \frac{4!(3n^{3}+(-1)^{n}n+3)}{n(n-1)(n-2)(n-3)}$. Hence $\lim_{n}c_{n}\leq 0$. But $c_{n}\geq 0$. Hence $\lim_{n}c_{n}=0$.

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  • $\begingroup$ This is basically a rewording of what it means to converge to $1$. I do not see this answer as very helpful. $\endgroup$ – Jason May 12 '17 at 16:20

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