極限值limit算法怎麼算 求解?

as x → 0

1/ln(1+x) - 1/x
= [ x - ln(1+x) ] / [ x*ln(1+x) ]
→ [ 1 - 1/(1+x) ] / [ ln(1+x) + x/(1+x) ] , by the L'Hopital's Rule
= [ x/(1+x) ] / [ ln(1+x) + x/(1+x) ]
= x / [ (1+x)*ln(1+x) + x ] , 因為 1+x → 1 ≠ 0 , 分子分母同乘有界的非0值 1+x , 結果不變
→ 1 / [ ln(1+x) + (1+x)/(1+x) + 1 ] , L'Hopital's Rule
→ 1 / ( 0 + 1 + 1 )
= 1/2 ..... Ans