微積分極限值的問題
回答 (3)
2012-07-01 5:12 am
✔ 最佳答案
求極限值lim(x->0)_Sin(3x)/xSollim(x->0)_Sin(3x)/x 0/0 type=lim3Cos(3x)/1=3orSinx=x－x^3/3!+x^5/5!－x^7/7!+…Sin(3x)=3x－(3x)^3/3!+(3x)^5/5!－(3x)^7/7!+…Sin(3x)/x=3－x^2*3^3/3!+x^4*3^5/5!－x^6*3^7/7!+…lim(x->0)_Sin(3x)/x=3
2012-07-01 9:42 am
lim(x→0) sin 3x/x
=lim(x→0) (sin3x)'/(x)'
=lim(x→0) 3cos(3x)/1
=3cos(0)/1
=3
=lim(x→0) (sin3x)'/(x)'
=lim(x→0) 3cos(3x)/1
=3cos(0)/1
=3
2012-07-01 8:07 am
lim(x→0) sin 3x/x
= lim(3x→0) sin 3x/x
= lim(3x→0) 3 sin 3x/3x
= 3 (as lim(3x→0) sin 3x/3x = 1)
= lim(3x→0) sin 3x/x
= lim(3x→0) 3 sin 3x/3x
= 3 (as lim(3x→0) sin 3x/3x = 1)
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