lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
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lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放lim(x-0+)x
lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
lim(x-0+)x∧(x);用洛必达法则求极限.
x的x次放
lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
lim(x-0+) x^x = lim(x-0+) e^ (x lnx) = e^{ lim(x-0+) x lnx }
lim(x-0+) x lnx = lim(x-0+) lnx / (1/x)
= lim(x-0+) (1/x) / (-1/x²) = lim(x-0+) -x = 0
=> lim(x-0+) x^x = e^0 = 1
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