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第一個無極限 第二個為0 第一個lim x->0 sin(1/x)= lim t->無窮sin(t) 若極限存在為a不等於0,即當t>t0之後sin(t)=a,則sin(t+pi)=-a不等於a,所以極限不存在 若極限為0,取t=t0+pi/2,sin(t)=1,所以a不為零 第二個因為|sin(1/x)|
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