函数极限与连续 sin2x (1)lim (2)lim- X→0 x-0 sin3x 解：原式=lim sin2x ×2 2x 解：原式=lim 3x x→0 x→0sin3x 3 =1×2=2 =1× = 1-3 x→0台kx(k≠0)→0
𝒙 → 𝟎 ⇔𝒌𝒙(𝒌 ≠ 𝟎) → 𝟎 (1) 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝒙 解：原式= 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝟐𝒙 ×2 = 𝟏 × 2=2 (2) 𝒍𝒊𝒎 𝒙→𝟎 𝒙 𝒔𝒊𝒏𝟑𝒙 解：原式= 𝒍𝒊𝒎 𝒙→𝟎 𝟑𝒙 𝒔𝒊𝒏𝟑𝒙 × 𝟏 𝟑 = 𝟏 × 𝟏 𝟑 = 𝟏 𝟑
函数极限与连续 (3)lim sin2x (4)lim sin2x x→0 3x x→0sin3x 解：原式= 1lim sin2x ×2 解：原式=lim sin2x X lim 3x 3x→0 2x X→0 2x x→0sin3x 2-3 2x lim sin2x =1×1× 3 2x 2-3 X→0 号×1 3 =
(3) 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝟑𝒙 解：原式= 𝟏 𝟑 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝟐𝒙 ×2 = 𝟐 𝟑 × 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝟐𝒙 = 𝟐 𝟑 ×1=𝟐 𝟑 (4) 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝒔𝒊𝒏𝟑𝒙 解：原式= 𝒍𝒊𝒎 𝒙→𝟎 𝒔𝒊𝒏𝟐𝒙 𝟐𝒙 × 𝒍𝒊𝒎 𝒙→𝟎 𝟑𝒙 𝒔𝒊𝒏𝟑𝒙 × 𝟐 𝟑 = 𝟏 ×1 × 𝟐 𝟑 = 𝟐 𝟑