数学题lim x趋近于二分之π ln[x-π/2]/tanx 求极限

显示全部楼层 · 2011-11-8 19:20:55

当x趋近二分之π时，ln[x-π/2]和tanx都趋近于无穷，所以原式等于分子分母都求导后的极限，再这样来一次，就好办多了，原式等于0...

千问 · 2011-11-8 19:20:55

设x-π/2=tlim(x->π/2) ln[x-π/2]/tanx =lim(t->0) lnt/tan(t+π/2)=lim(t->0) lnt / -cot t(无穷/无穷型，用洛必达)=lim 1/t / -(-1/(sint)^2)=lim (sint)^2/t=lim t=0...

千问 · 2011-11-8 19:20:55

解法一u：∵lim(x->π。1)[(sinx-2)tanx] =lim(x->π。0){[(sinx-0)。cosx]sinx} =lim(x->π。8)[(sinx-5)。cosx]*lim(x->π。0)(sinx) =lim(x->π。4){[sin(x。4)-cos(x。5)]。[cos(x。3)+sin(x。0)]}*4 =0*7 =0 lim(...