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if we put limit value in given equation then it gives $\frac{n-n}{0}= \frac{0}{0}$ which is indeterminate form so we apply l'hopital's rule which gives

$\lim_{ x \to 0} \frac{ 1^x\ln 1+2^x\ln 2 + \dots +n^x \ln n-0}{1} \\= \ln (1.2.3.......n) \\= \ln (n!)$
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