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3)

lim x-->0 (sinx-xcosx)/tanx =?
0/0 indefinite
L'Hopital rule:
lim x-->0 (cosx-(cosx-xsinx))/(1+tan^x)

lim x-->0 xsinx /(1+tan^x)
=0/1=0

2007-10-19 07:37:27 · answer #1 · answered by iyiogrenci 6 · 0 0

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