Question

求极限：（洛必达法则）

Lim(x→0+) lnx*ln(x+1)
Lim(x→1) (1-x)tan(xπ/2)
Lim(x→0) (sinx)^(x)
Lim(x→0) (1+sinx)^(1/x)
Lim(x→π/2+) (tanx)^(cosx)

Answer 1

将上述表达式按如下化简，然后运用洛比达法则对分子分母求导即可，后续步骤及结果略……
Lim(x→0+) lnx*ln(x+1) = Lim(x→0+) ln(x+1) / (1/lnx)
Lim(x→1) (1-x)tan(xπ/2) = Lim(x→1) tan(xπ/2)/ (1/(1-x))
Lim(x→0) (sinx)^(x) = Lim(x→0) e^[xln(sinx)] = Lim(x→0) e^[ln(sinx)/ (1/x)]
Lim(x→0) (1+sinx)^(1/x) = Lim(x→0) e^[ (ln(1+sinx) )/ x]
Lim(x→π/2+) (tanx)^(cosx) = Lim(x→π/2+) e^[ (ln(tanx) ) /(1/cosx) ]