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Thm:lim(x→0)sinx=0,lim(x→0)cosx=1 Hence sinx and cosx ...

3x is cont. at 0 and cos is cont. at 0. c→f(c)→g(f(c)) ... ⇒lim(x→0)cos(3x)=cos0=1 函數連續，該點極限值等於該點函數值 ... A:無限多個。

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