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$$\lim\limits_{n \to \infty}\frac{n\cos(n^4)}{n^3+\ln (n)}$$

I'm having trouble on how to approach finding this limit, but I do know the limit is 0 since $\cos(n^4)$ is just between -1 and 1 and $\ln(n)$ grows very slowly so for $n$ sufficiently large it will just be $\frac{n}{n^3}$ or $\frac{1}{n^2}$ and that converges to 0.

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  • $\begingroup$ The corresponding infinite series is absolutely converges. $\endgroup$ – mrs Nov 26 '16 at 11:54
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$\dfrac{n\cos(n^4)}{n^3+\ln n}=\dfrac{n\cos(n^4)}{n^3}\dfrac{1}{1+\frac{\ln n}{n^3}}\sim \dfrac{cos(n^4)}{n^2}\to 0$.

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Use the squeeze theorem:

$$0\xleftarrow[\infty\leftarrow n]{}\frac{-1}{n^2}=\frac{-n}{2n^3}\le\frac{n\cos n^4}{n^3+\log n}\le\frac n{n^3}=\frac1{n^2}\xrightarrow[n\to\infty]{}0$$

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