Re: [微積] L'hospitla rule
※ 引述《ytyt5239 (ytyt)》之銘言：
: 1.(跟定理沒關係)find the vertical and horizantal asymptotes.
: f(x)=ln(1-ln x)
: 我是想 t=ln x →無限 as x→0+
: lim x→0:ln(1-ln x)=lim t→無限:ln(1-t)
: 然後就不知怎辦了～
(1)1-ln x >0 x<e
x→e+ f(x)→-∞
(2)x→0+ f(x)→+∞
: 2.evalute lim x→無限 [x-x^2‧ln(1+x/x)]
: 我只想只ln as the same factor ,and then i dont know how to resolve this
: question , help me out!!!
lim(x→∞) [x-x^2 ln(1+1/x)]
= lim(x→∞) x[1-xln(1+1/x)] let t=1/x
= lim(t→0+) [1-(ln(1+t))/t]/t
= lim(t→0+) [t-ln(1+t)]/t^2 L'Hospital's Rule (0/0)
= lim(t→0+) [1-1/(1+t)]/2t
= lim(t→0+) [1+t-1]/2t(t+1) =0.5
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