今利用连续性求极限举例 例5求im x->0 2 解limy+x (√1+x2-1)(√1+x2+1) Im x->0 x(√1+x2+1) 0 x>0√1+x2+12 0 例6求m8(1+x) x->0 xX 解 lim logo(l+x) lim log (+x)x=loge x→ 0 xX x->0 页返回 页结束铃首页 上页 返回 下页 结束 铃 例 8 求 x x a x log (1 ) lim 0 + → 例6  例 例 7 5 求 x x x 1 1 lim 2 0 + - →  解 x x x 1 1 lim 2 0 + - → ( 1 1) ( 1 1)( 1 1) lim 2 2 2 0 + + + - + + = → x x x x x 0 2 0 1 1 lim 0 2 = = + + = → x x x  解 x x a x log (1 ) lim 0 + → x a x x 1 0 = lim log (1+ ) → a ea ln 1 =log =  解 解 解 x x x 1 1 lim 2 0 + - → ( 1 1) ( 1 1)( 1 1) lim 2 2 2 0 + + + - + + = → x x x x x 0 2 0 1 1 lim 0 2 = = + + = → x x x  解 x x a x log (1 ) lim 0 + → x a x x 1 0 = lim log (1+ ) → a ea ln 1 解 =log =  x x a x log (1 ) lim 0 + → x a x x 1 0 = lim log (1+ ) → a ea ln 1 =log =  下页 ❖利用连续性求极限举例