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Question Number 62332 by tanmay last updated on 19/Jun/19

Answered by tanmay last updated on 20/Jun/19

$$\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{\sqrt{2}\times n^{n-\frac{1}{2}}}{n^{n+\frac{1}{2}}\times \sqrt{2\pi }\times e^{-n}}\times {\left\{\frac{{(2\times n^{\frac{1}{n}}-1)}^n}{n^2}\right\}}^{\frac{n(n-\frac{1}{2})}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{{(2\times n^{\frac{1}{n}}-1)}^n}{n^2}\right\}}^{\frac{n(n-\frac{1}{2})}{{ln}^{2}n}}\right]$$$$now$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{{(2n^{\frac{1}{n}}-1)}^{n^2}}{n^{2n}}\right\}}^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{n^n{(2-\frac{1}{n^{\frac{1}{n}}})}^{n^2}}{n^n\times n^n}\right\}}^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{2^{n^2}{(1-\frac{1}{2n^{\frac{1}{n}}})}^{n^2}}{n^n}\right\}}^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{2^{n^2}}{n^n}\times {(1-\frac{1}{2n^{\frac{1}{n}}})}^{2n^{\frac{1}{n}}\times \frac{n^2}{2n^{\frac{1}{n}}}}\right\}}^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times {\left\{\frac{2^{n^2}}{n^n}\times e^{-\frac{n^2}{2n^{\frac{1}{n}}}}\right\}}^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$=\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{e^n}{n}\times 2^{n^2\times \frac{n-\frac{1}{2}}{{ln}^{2}n}}\times \frac{1}{n^{\frac{n-\frac{1}{2}}{{ln}^{2}n}}}\times {\left(e^{\frac{-1}{2}}\right)}^{\frac{n^2}{n^{\frac{1}{n}}}\times \frac{n-\frac{1}{2}}{{ln}^{2}n}}\right]$$$$\frac{1}{\sqrt{\pi }}\underset{n\to \mathrm{\infty }}{\mathrm{li}m}\mathit{w}\mathit{a}\mathit{i}\mathit{t}$$

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