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lim(x→0)sinx/x=1 這是兩個重要極限之一.屬於 0/0 型極限,也可以使用洛必達法則求出. lim(x→0)sinx/x=lim(x→0)cosx/1=1/1=1 lim(x->∞) sinx/x = 0 分子 sinx
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