⑩天掌 Teaching Plan on Advanced Mathematics 例4求m+x2-1 x→>0 解原式=lim (√1+x2-1)(1+x2+1) x-0 x(√1+x2+1) =0 x0√1+x2+1 例5求lm og,(1+x) x→>0 解原式= lim log,(+x) =log lim(1+x)=lne=1Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例4 . 1 1 lim 2 0 x x x + − → 求 解 ( 1 1) ( 1 1)( 1 1) lim 2 2 2 0 + + + − + + = → x x x x 原式 x 1 1 lim 2 0 + + = → x x x 2 0 = = 0. 例5 . x log (1 x) lim a x 0 + 求 → x 1 a x 0 = lim log (1+ x) 解 原式 → (1 x) ] = 1. log [lim x 1 x 0 = a + → = lne