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標題:
a-maths~!!!!!
發問:
http://www.photo-host.org/img/467859screenhunter_01_jul._01_17.53.gif
最佳解答:
let k= (4+x)/(4-x) 4k - kx= 4 + x 4(k-1 ) = x( k+1) x/4 = (k-1)/ (k+1) = 1 - 2/( k+1) so, lim( x->4) x /4 = 1 = 1- 2/( k+1) as there is no solution of k , lim( x->4) (4+x)/(4-x) is no defined . 2008-07-01 19:22:59 補充： 可用上極限and下極限來做, check下 lim(x-> +4 ) f(x) = lim(x-> -4) f(x) ???
Put x=4 into the fraction, which is not in the form of 0/0, infinity/0, or 0/infinity, which implies that the limit does not exists
a-maths~!!!!!
發問:
http://www.photo-host.org/img/467859screenhunter_01_jul._01_17.53.gif
最佳解答:
let k= (4+x)/(4-x) 4k - kx= 4 + x 4(k-1 ) = x( k+1) x/4 = (k-1)/ (k+1) = 1 - 2/( k+1) so, lim( x->4) x /4 = 1 = 1- 2/( k+1) as there is no solution of k , lim( x->4) (4+x)/(4-x) is no defined . 2008-07-01 19:22:59 補充： 可用上極限and下極限來做, check下 lim(x-> +4 ) f(x) = lim(x-> -4) f(x) ???
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其他解答:Put x=4 into the fraction, which is not in the form of 0/0, infinity/0, or 0/infinity, which implies that the limit does not exists
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