Question

习题1.3.1lima[x]0的左极限和右极限分别都是怎么算的？过程详细

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Answer 1

lim(x->0+) { ln[1+e^(2/x)] /ln[(1+e^(1/x)] + a[x] }
=lim(x->0+) ln[1+e^(2/x)] /ln[(1+e^(1/x)] (∞/∞ 分子分母分别求导)
=lim(x->0+) { -(2/x^2)/ [1+e^(2/x)] }/{ (-1/x^2)/[(1+e^(1/x)] }
=lim(x->0+) -2[(1+e^(1/x)]/ [1+e^(2/x)]
分子分母同时除以 e^(2/x)
=lim(x->0+) -2[(1/e^(2/x)+1/e^(1/x)]/ [1/e^(2/x) +1]
=0
lim(x->0-) { ln[1+e^(2/x)] /ln[(1+e^(1/x)] + a[x] }
=-a +lim(x->0-) ln[1+e^(2/x)] /ln[(1+e^(1/x)] (0/0 分子分母分别求导)
=-a +lim(x->0-) { -(2/x^2)/ [1+e^(2/x)] }/{ (-1/x^2)/[(1+e^(1/x)] }
=-a +lim(x->0-) -2[(1+e^(1/x)]/ [1+e^(2/x)]
=-a - 2(1-0)/(1+0)
=-a -2
f(0+)=f(0-)
-a-2 =0
a=-2