f(x)=(1-a^(1/x))/(1+a^(1/x))(a>1),求lim(x→0)f(x)f(x)在x=0处没有定义,而lim(x→0^+)f(x)=lim(x→0^+)(a^(-1/x)-1)/(a^(-1/x)+1)=-1{为何与原式不同}lim(x→0^-)f(x)=lim(x→0^-)(1-a^(1/x))/(1+a^(1/x))=1因为lim(x→0^+
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![f(x)=(1-a^(1/x))/(1+a^(1/x))(a>1),求lim(x→0)f(x)f(x)在x=0处没有定义,而lim(x→0^+)f(x)=lim(x→0^+)(a^(-1/x)-1)/(a^(-1/x)+1)=-1{为何与原式不同}lim(x→0^-)f(x)=lim(x→0^-)(1-a^(1/x))/(1+a^(1/x))=1因为lim(x→0^+](/uploads/image/z/1076872-40-2.jpg?t=f%28x%29%3D%281-a%5E%281%2Fx%29%29%2F%281%2Ba%5E%281%2Fx%29%29%28a%3E1%29%2C%E6%B1%82lim%28x%E2%86%920%29f%28x%29f%28x%29%E5%9C%A8x%3D0%E5%A4%84%E6%B2%A1%E6%9C%89%E5%AE%9A%E4%B9%89%2C%E8%80%8Clim%EF%BC%88x%E2%86%920%5E%2B%EF%BC%89f%28x%29%3Dlim%EF%BC%88x%E2%86%920%5E%2B%EF%BC%89%28a%5E%28-1%2Fx%29-1%29%2F%28a%5E%28-1%2Fx%29%2B1%29%3D-1%7B%E4%B8%BA%E4%BD%95%E4%B8%8E%E5%8E%9F%E5%BC%8F%E4%B8%8D%E5%90%8C%7Dlim%EF%BC%88x%E2%86%920%5E-%EF%BC%89f%28x%29%3Dlim%EF%BC%88x%E2%86%920%5E-%EF%BC%89%281-a%5E%281%2Fx%29%29%2F%281%2Ba%5E%281%2Fx%29%29%3D1%E5%9B%A0%E4%B8%BAlim%EF%BC%88x%E2%86%920%5E%2B)
f(x)=(1-a^(1/x))/(1+a^(1/x))(a>1),求lim(x→0)f(x)f(x)在x=0处没有定义,而lim(x→0^+)f(x)=lim(x→0^+)(a^(-1/x)-1)/(a^(-1/x)+1)=-1{为何与原式不同}lim(x→0^-)f(x)=lim(x→0^-)(1-a^(1/x))/(1+a^(1/x))=1因为lim(x→0^+
f(x)=(1-a^(1/x))/(1+a^(1/x))(a>1),求lim(x→0)f(x)
f(x)在x=0处没有定义,而
lim(x→0^+)f(x)=lim(x→0^+)(a^(-1/x)-1)/(a^(-1/x)+1)=-1
{为何与原式不同}
lim(x→0^-)f(x)=lim(x→0^-)(1-a^(1/x))/(1+a^(1/x))=1
因为lim(x→0^+)f(x)≠lim(x→0^-)f(x),故lim(x→0)f(x)不存在
f(x)=(1-a^(1/x))/(1+a^(1/x))(a>1),求lim(x→0)f(x)f(x)在x=0处没有定义,而lim(x→0^+)f(x)=lim(x→0^+)(a^(-1/x)-1)/(a^(-1/x)+1)=-1{为何与原式不同}lim(x→0^-)f(x)=lim(x→0^-)(1-a^(1/x))/(1+a^(1/x))=1因为lim(x→0^+
分子分母同除a^(1/x),原式变为f(x)=(a^x-1)/(a^x+1),
f(x)在x=0时为∞/∞形,所以用洛必达法则,
将分子分母同时求导,即
lim(x→0)f(x)=lim(x→0)(a^x-1)的导数/(a^x+1)的导数=1
是原式通过上下同除以一个a^(1/x)得到的化简式
因为a^(-1/x)---->0,因为左右极限不同,所以x--->0处不存在极限