例19.已知limx+10012+00 求c x→∝ 解 Iimx+1001、2x+00 x→0 5 1006 1006 x+1001 lim(1+ 100x35lim( x→0 x→0 x-5 lim⊥10 006 x-52012x 1006 2012 c=201215 例19. 已知 1001 2 2000 lim( ) , 5 x c x x e x + → + = − 求 c. 2 2000 5 10062 1006 5 2000 1001 lim( ) 5 1006 1001 lim(1 ) lim( ) 5 5 x x x x x x x x x x x x + →  −  − → → + − + = +  − − 解 2012  = c 5 2012 1006 1006 5 2012 = lim(1 ) 5 x x x c x e e x −  − →  + = = −