求极限:(x→1) lim(1-X)tg(π/2x)tg(π/2 x)等价于pai/2x :(x→1) lim(1-X)tg(π/2x) =lim(x->1)(1-x)pai/2x =lim(x->1)pai/2x-pai/2 =pai/2-pai/2 =0上面的做法错在哪里?(x→1) lim(1-X)tg(π/2 x) =lim(1-X)*cot(π/2-π/2 x) =lim(1-X)/tan(π
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