微積分 20點

1. lim(x->0)_x[－2/x]
Sol
(1) lim(x->0+)_x[－2/x]
=lim(y->∞)_(1/y)[－2y]
=lim(y->∞)_－2y/y
=－2
(2) lim(x->0-)_x[－2/x]
=lim(y->-∞)_(1/y)[－2y]
=lim(y->∞)_－2y/y
=－2
So
lim(x->0)_x[－2/x]=－2

2. lim(x->0)_[1/(xTanx)](1－Cosx)
=lim(x->0)_(1－Cosx)//(xTanx).. 0/0 type
=lim(x->0)_Sinx/(xSec^2 x+Tanx) 0/0type
=lim(x->0)_Cosx/(2xSec^2 xTanx+Sec^2 x+Sec^2 x)
=1/(2*0+1+1)
=1/2

3. f(x)=x^(1/3)*｜2x－7｜求f'(3)=?
Sol
f(3)=3^(1/3)
f’(3)=lim(x->3)_[f(x)－f(3)]/(x－3)
=lim(x->3)_[x^(1/3)*(7－2x)－3^(1/3)]/(x－3) 0/0 type
=lim(x->3)_[－2x^(1/3)+(1/3)*x^(－2/3)*(7－2x)]/1
=－2*3^(1/3)+(1/3)*3^(－2/3)*1
=[－2*9*3^(1/3)+(1/3)*9*3^(－2/3)]/9
=[－18*3^(1/3)+3^(1/3)]/9
=－17*3^(1/3)/9

4.lim(x->π/4-)_[√(1－Sin2x)/(4x－π)]
=lim(x->π/4-)_｜Cosx－Sinx｜/(4x－π)
=lim(x->π/4-)_(Cosx－Sinx)/(4x－π) 0/0 type
=lim(x->π/4-)_(－Sinx－Cosx)/4
=－√2/4

有人居然回答得出來
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