※ 引述《shallow1112 (小扯)》之銘言：
: lim =(x-π/4)^2/(tanx-1)^2
: x->π/4
: 不好意思
: 麻煩大家了
(x - π/4)^2
原題= l i m ----------------------
x->π/4 (tanx)^2 - 2tanx + 1
[(cosx)^2]*[(x - π/4)^2]
= l i m -------------------------------------
x->π/4 [(sinx)^2] - 2sinxcosx + [(cosx)^2]
[(cosx)^2]*[(x - π/4)^2]
= l i m -----------------------------
x->π/4 1 - 2sinxcosx
令u = x - π/4 , 當 x -> π/4時 , u -> 0
(u^2)*{[√2/2](cosu - sinu)}^2
上式 = l i m ------------------------------
u ->0 1 - cos2u
(1/2)(u^2)*[(cosu - sinu)^2] [(2u)^2]/2
= l i m {----------------------------- * -------------}
u ->0 [(2u)^2]/2 1 - cos2u
= (1/4) * 1 = 1/4
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