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Question Number 132321 by Raxreedoroid last updated on 13/Feb/21

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{\cos \left(x\ln k\right)}{{\left(\ln k\right)}^n\sqrt{k}}=?$$$$\mathrm{where}k,n\in \mathbb{N},x\in \mathbb{R}$$

Answered by Dwaipayan Shikari last updated on 13/Feb/21

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{cos\left(log\left(k^x\right)\right)}{{\left(log\left(k\right)\right)}^n\sqrt{k}}=\frac{k^{xi}+k^{-xi}}{\sqrt{k}{\left(log\left(k\right)\right)}^n}=\frac{k^{xi-\frac{1}{2}}+k^{xi-\frac{1}{2}}}{{\left(log\left(k\right)\right)}^n}=\mathrm{\Phi }$$$$Ifk>e\mathrm{\Phi }=0$$$$k=e\mathrm{\Phi }=\frac{cos\left(log\left(k^x\right)\right)}{\sqrt{k}}$$$$k<e\mathrm{\Phi }\to \mathrm{\infty }$$

Commented by Raxreedoroid last updated on 13/Feb/21

$$\mathrm{sir}\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{always}0\mathrm{except}\mathrm{for}\mathrm{k}=1$$$$\mathrm{since}\cos \left(x\right)\mathrm{where}x\in \mathbb{R}\mathrm{is}\mathrm{between}[-1,1]$$$$\mathrm{and}\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{a\cdot b^x}=0$$$$\mathrm{so}\mathrm{in}\mathrm{that}\mathrm{case}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{\cos \left(\log \left({\mathrm{k}}^{\mathrm{x}}\right)\right)}{{\left(\log \left(\mathrm{k}\right)\right)}^{\mathrm{n}}\sqrt{k}}=\frac{[-1,1]}{\mathrm{incredibly}\mathrm{large}}=0,k\ne 1$$

Commented by Raxreedoroid last updated on 13/Feb/21

$$\mathrm{sorry}\mathrm{my}\mathrm{mistake}$$$$\ln \left(\mathrm{k}\right)\mathrm{where}\mathrm{k}>1\mathrm{is}\mathrm{between}(0,\mathrm{\infty })$$$$\mathrm{so}\mathrm{your}\mathrm{answer}\mathrm{was}\mathrm{right}$$

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