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Author: Tinku Tara

lim-n-k-1-n-1-1-k-1-n-1-k-1-n-1-k-1-n-

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Question Number 215748 by MrGaster last updated on 17/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} −\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}}} \\ $$ Answered by MrGaster last updated on 02/Feb/25…

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Question-215776

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Question Number 215776 by cherokeesay last updated on 17/Jan/25 Answered by A5T last updated on 18/Jan/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{be}\:\mathrm{r} \\ $$$$\mathrm{AC}\parallel\mathrm{OB}\Rightarrow\mathrm{CB}=\sqrt{\mathrm{r}^{\mathrm{2}} +\left(\mathrm{r}−\mathrm{x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2rx}} \\ $$$$\mathrm{Power}\:\mathrm{point}\:\mathrm{of}\:\mathrm{C}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}…

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1-1-1-x-2-1-x-2-1-1-x-2-y-2-1-1-y-2-x-2-y-2-z-2-5-2-dz-dy-dx-is-

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Question Number 215740 by universe last updated on 16/Jan/25 $$ \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} ^{\mathrm{1}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}}…

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Question-215739

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Question Number 215739 by universe last updated on 16/Jan/25 Answered by MrGaster last updated on 19/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{n}^{{k}} }{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{n}} } \\…

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lim-x-ln-9-2-1-2-x-

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Question Number 215718 by mathlove last updated on 16/Jan/25 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({ln}\frac{\mathrm{9}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} =? \\ $$$$ \\ $$ Answered by mehdee7396 last updated on 17/Jan/25 $${if}\:\:{x}\rightarrow+\infty\Rightarrow{lim}\left({ln}\frac{\mathrm{9}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} =+\infty…

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