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Calculate the limit:

Posted August 13th, 2010 by alpha
in

Evaluate the limits:
$ \mathop{\lim}\limits_{n \to \infty}{\frac{n^2-n+1}{n^2-n-1}}^{n(n-1)} $
$ \mathop{\lim}\limits_{x \to \infty}\frac{2\sqrt{x}+3\sqrt[3]{3}+4\sqrt[4]{4}.....+\sqrt[n]{n}}{\sqrt{2x-3}+\sqrt[3]{2x-3}+............\sqrt[n]{2x-3}} $

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Comments

answer

On August 14th, 2010 Structure says:
$$\mathop{\lim}\limits_{n \to \infty}({\frac{n^2-n+1}{n^2-n-1}})^{n(n-1)} =\mathop{\lim}\limits_{n \to \infty}(1+{\frac{2}{n^2-n-1}})^{n(n-1)}=\mathop{\lim}\limits_{n \to \infty}(1+{\frac{2}{n^2-n-1}})^{\frac{n^2-n-1}{2}\mathop{\lim}\limits_{n \to \infty}\frac{2n(n-1)}{n^2-n-1}}=e^2$$
$$\mathop{\lim}\limits_{x \to \infty}\frac{2\sqrt{x}+3\sqrt[3]{3}+4\sqrt[4]{4}.....+\sqrt[n]{n}}{\sqrt{2x-3}+\sqrt[3]{2x-3}+............\sqrt[n]{2x-3}}=\mathop{\lim}\limits_{x \to \infty}\frac{\frac{2\sqrt{x}}{\sqrt x}+\frac{3\sqrt[3]{3}}{\sqrt x}+\frac{4\sqrt[4]{4}}{\sqrt x}.....+\frac{\sqrt[n]{n}}{\sqrt x}}{\frac{\sqrt{2x-3}}{\sqrt x}+\frac{\sqrt[3]{2x-3}}{\sqrt x}+............\frac{\sqrt[n]{2x-3}}{\sqrt x}}=\sqrt 2$$

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