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Question Number 159405 by qaz last updated on 16/Nov/21

$$\underset{\mathrm{x}\to +0}{\lim }{\left(\frac{\underset{\mathrm{k}=1}{\sum ^{2021}}{\mathrm{k}}^{\mathrm{x}}}{2021}\right)}^{\frac{1}{\mathrm{x}}}=?$$

Answered by mindispower last updated on 16/Nov/21

$$=\underset{x\to 0}{\lim }{e}^{\frac{ln\left(\frac{\mathrm{\Sigma }{k}^x}{2021}\right)}{x}}$$$$=e^{\frac{ln\left(\underset{k=1}{\prod ^{2021}}k\right)}{2021}}={(2021!)}^{\frac{1}{2021}}$$

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