⑩天掌 Teaching Plan on Advanced Mathematics 例6求 x→0 解令a-1=t, 则x=lg(1+t 当x→0时,t→0. 原式=lim log (1+t) =lim t→0 log (1+t) naTianjin Polytechnic University Teaching Plan on Advanced Mathematics 例6 . 1 lim 0 x a x x − 求 → 解 a 1 t, x 令 − = x log (1 t), 则 = a + log (1 ) lim 0 t t a y + = 原式 → 当x →0时,t →0. t a t t 1 0 log (1 ) 1 lim + = → = lna