※ 引述《znmkhxrw (QQ)》之銘言：
: if f(∞)=0,g(∞)=0
: let x=1/u
: lim_{x→∞} f(x)/g(x)
: = lim_{u→0+} f(1/u)/g(1/u)
You can take f(1/u)=p(u) and g(1/u)=q(u)
when u→0+, then p(u)→0, q(u)→0
Use ORDINARY HOSPITAL's RULE
lim_{u→0+} f(1/u)/g(1/u)
= lim_{u→0+} p(u)/q(u)
= lim_{u→0+} p'(u)/q'(u)
= lim_{u→0+} [f'(1/u)*(-1/u^2)]/[g'(1/u)*(-1/u^2)]
= lim_{u→0+} f'(1/u)/g'(1/u)
= lim_{x→∞} f'(x)/g'(x)
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ORDINARY HOSPITAL's RULE
when u→0+, then p(u)→0, q(u)→0
We can define
P(u)=p(u) when u>0
Q(u)=q(u) when u>0
and DEFINE P(0)=Q(0)=0
then
lim_{u→0+} p(u)/q(u)
=lim_{u→0+} P(u)/Q(u)
=lim_{u→0+} [P(u)/u]/[Q(u)/u]
=lim_{u→0+} [P(u)/u] / lim_{u→0+} [Q(u)/u]
=P'(0+)/Q'(0+)
=p'(0+)/q'(0+)
=lim_{u→0+} p'(u)/q'(u)
In fact, I don't like analysis.
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◆ From: 112.104.142.5
※ 編輯: JohnMash 來自: 112.104.142.5 (07/06 12:48)