lim x tends to 0 ((1^x+2^x+3^x+....+n^x)/n)^a/x =?

exponential limits
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1^0=1, 2^0=1, 3^0=1, ....n^0=1 so yu hav n/n=1 then yu hav [1]^a & that=1 but then yu hav /x, so me ges the hole dang thang is 1/x
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