1+2+…n (3)lim( x→0 n 解(3) …+n lim-In( 原式 =pro lim +...+n =ex)0(x) 11In1+2xIn2+…nlnn lim nn. =e>01 1+ =e n=nnx x x x x n n 1 0 ) 1 2 (3) lim( + + → ) 1 ln( 1 lim 0 (3) n n x x x x e + + = →  原 式 解 [ln(1 ) ln( )] ( ) 1 lim 0 + + −  →  = n n x x x x e  x x x x x x n n n e   + + +  → = 1 1 ln1 2 ln 2 ln 1 1 lim 0 n n n e n! ln ! = =