evaluate the following limits
1. lim x->2 (x-2) / (√(x^2-4))
2. lim x->5 (x-5)/ (√(x+4)-3)
limits
2007-03-05 6:41 am
回答 (3)
2007-03-06 12:59 am
2007-03-05 7:19 pm
lim x->2 (x-2) / (√(x^2-4))
=lim x->2 (x-2) / (√(x-2)(x+2)
=lim x->2 √(x-2)^2/(x+2)(x-2)
=lim x->2 √(x-2/x+2)
=√0/4
=√0
=0
lim x->5 (x-5)/ (√(x+4)-3)
=limx->5 (x-5)(√(x+4)+3)/(√(x+4)-3)(√(x+4)+3)
=limx->5 (x-5)(√(x+4)+3)/(x+4+9)
=limx->5 (x-5)(√(x+4)+3)/(x+13)
=(5-5)(√(5+4)+3)/(5+13)
=0/18
=0
2007-03-05 11:21:46 補充：
lim x->5 (x-5)/ (√(x+4)-3)=limx->5 (x-5)(√(x+4)+3)/(x+4-9)=limx->5(x-5)(√(x+4)+3)/(x-5)=limx->5 (√(x+4)+3)=(√(5+4)+3)=6
=lim x->2 (x-2) / (√(x-2)(x+2)
=lim x->2 √(x-2)^2/(x+2)(x-2)
=lim x->2 √(x-2/x+2)
=√0/4
=√0
=0
lim x->5 (x-5)/ (√(x+4)-3)
=limx->5 (x-5)(√(x+4)+3)/(√(x+4)-3)(√(x+4)+3)
=limx->5 (x-5)(√(x+4)+3)/(x+4+9)
=limx->5 (x-5)(√(x+4)+3)/(x+13)
=(5-5)(√(5+4)+3)/(5+13)
=0/18
=0
2007-03-05 11:21:46 補充：
lim x->5 (x-5)/ (√(x+4)-3)=limx->5 (x-5)(√(x+4)+3)/(x+4-9)=limx->5(x-5)(√(x+4)+3)/(x-5)=limx->5 (√(x+4)+3)=(√(5+4)+3)=6
2007-03-05 7:00 am
1.
lim x->2 (x-2) / (√(x^2-4))
= lim x-> 2 (x - 2) / √(x + 2)(x - 2)
= lim x-> 2 √ {[(x - 2)]^2 / (x + 2)(x - 2)}
= limx->2 √[(x - 2) / (x + 2)]
= 0
2.
lim x->5 (x-5)/ (√(x+4)-3)
= lim x->5 (x - 5)(√(x + 4) + 3) / (x + 4 - 9)
= lim x->5 (x - 5)(√(x + 4) + 3) / (x - 5)
= lim x->5 (√(x + 4) + 3)
= 3 + 3
= 6
lim x->2 (x-2) / (√(x^2-4))
= lim x-> 2 (x - 2) / √(x + 2)(x - 2)
= lim x-> 2 √ {[(x - 2)]^2 / (x + 2)(x - 2)}
= limx->2 √[(x - 2) / (x + 2)]
= 0
2.
lim x->5 (x-5)/ (√(x+4)-3)
= lim x->5 (x - 5)(√(x + 4) + 3) / (x + 4 - 9)
= lim x->5 (x - 5)(√(x + 4) + 3) / (x - 5)
= lim x->5 (√(x + 4) + 3)
= 3 + 3
= 6
收錄日期: 2021-04-12 21:20:39
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