lim-x-0-1-x-2-x-n-x-n-1-x-

Question Number 164705 by Kayela last updated on 20/Jan/22

\lim_{x \to 0} {(\frac{1^{x} + 2^{x} + \dots + n^{x}}{n})}^{\frac{1}{x}}

Answered by Berlindo last updated on 20/Jan/22

= {[\lim_{x \to 0} {(1 + \frac{1^{x} + 2^{x} + \dots + n^{x} - n}{n})}^{\frac{n}{1^{x} + 2^{x} + \dots + n^{x} - n}}]}^{\lim_{x \to 0} \frac{1^{x} + 2^{x} + \dots + n^{x} - n}{x} \cdot \frac{1}{n}}

= e^{\lim_{x \to 0} \frac{(1^{x} - 1) + (2^{x} - 1) + \dots + (n^{x} - 1)}{x} \cdot \frac{1}{n}}

= e^{[\lim_{x \to 0} \frac{1^{x} - 1}{x} + \lim_{x \to 0} \frac{2^{x} - 1}{x} + \dots + \lim_{x \to 0} \frac{n^{x} - 1}{x}] \cdot \frac{1}{n}}

= e^{(\ln 1 + \ln 2 + \dots + \ln n) \cdot \frac{1}{n}} = e^{\frac{1}{n} \ln (n!)} = e^{\ln (\sqrt[n]{n!})}

= \sqrt[n]{n!}

∴ \lim_{x \to 0} {(\frac{1^{x} + 2^{x} + \dots + n^{x}}{n})}^{\frac{1}{x}} = \sqrt[n]{n!}

★ Berlindo Ndala ★

Answered by puissant last updated on 21/Jan/22

= \lim_{x \to 0} {(\frac{e^{x l n 1} + e^{x l n 2} + \dots + e^{x l n n}}{n})}^{\frac{1}{x}}

= \lim_{x \to 0} {(\frac{(1 + x l n 1) + (1 + x l n 2) + \dots + (1 + x l n n)}{n})}^{\frac{1}{x}}

= \lim_{x \to 0} {(1 + \frac{x l n (n!)}{n})}^{\frac{1}{x}}

= \lim_{x \to 0} e^{\frac{1}{x} l n (1 + \frac{x l n (n!)}{n})} = \lim_{x \to 0} e^{\frac{1}{x} (\frac{x l n (n!)}{n})}

= e^{\frac{l n (n!)}{n}} = \sqrt[n]{n!} . .

\dots \dots \dots \dots L e p u i s s a n t \dots \dots \dots . .

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