arctan 例3求mm 2 解原式 1+x li x→+ x→)+∞1+x In sin ax 例4求lm x→y0 In sin bx 解原式=lim a coax sin bx cosx =lim x→0 bosox. SIn(r x→0 cosar Economic-mathematics 27-13 Wednesday, February 24, 2021Economic-mathematics 27- 13 Wednesday, February 24, 2021 例3 解 . 1 arctan 2 lim x x x − →+   求 2 2 1 1 1 lim x x x − + − = →+  原式 2 2 1 lim x x x + = →+ = 1. 例4 解 . lnsin lnsin lim 0 bx ax x→ 求 b bx ax a ax bx x cos sin cos sin lim 0   = → 原式 = 1. ) 0 0 ( ( )   ax bx x cos cos lim →0 =