Morrison Clark
02/08/2023 · High School

\( \operatorname { Lim } _ { ( x , y ) \rightarrow ( 1,1 ) } \frac { x y - x - y + 1 } { x ^ { 2 } + y ^ { 2 } - 2 x - 2 y + 2 } \)

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为了求解极限问题 \[ \operatorname { Lim } _ { ( x , y ) \rightarrow ( 1,1 ) } \frac { x y - x - y + 1 } { x ^ { 2 } + y ^ { 2 } - 2 x - 2 y + 2 }, \] 我们可以按照以下步骤进行: ### 步骤 1:直接代入 首先,我们尝试直接代入 \( x = 1 \) 和 \( y = 1 \) 到极限表达式中: \[ \operatorname { Lim } _ { ( x , y ) \rightarrow ( 1,1 ) } \frac { 1 \cdot 1 - 1 - 1 + 1 } { 1^2 + 1^2 - 2 \cdot 1 - 2 \cdot 1 + 2 } = \frac { 1 - 1 - 1 + 1 } { 1 + 1 - 2 - 2 + 2 } = \frac { 0 } { 0 }, \] 得到一个不定式 \( \frac{0}{0} \),因此我们需要进一步分析。 ### 步骤 2:使用极坐标变换 为了简化问题,我们可以使用极坐标变换。设 \( x = r \cos \theta \) 和 \( y = r \sin \theta \),当 \( (x, y) \rightarrow (1, 1) \) 时,\( r \rightarrow 1 \) 且 \( \theta \) 可以取任意值。 将 \( x \) 和 \( y \) 代入原极限表达式中: \[ \operatorname { Lim } _ { r \rightarrow 1 } \frac { r^2 \cos \theta \sin \theta - r \cos \theta - r \sin \theta + 1 } { r^2 - 2 r \cos \theta - 2 r \sin \theta + 2 }. \] ### 步骤 3:简化分子和分母 分子可以简化为: \[ r^2 \cos \theta \sin \theta - r \cos \theta - r \sin \theta + 1. \] 分母可以简化为: \[ r^2 - 2 r \cos \theta - 2 r \sin \theta + 2. \] ### 步骤 4:分析极限 当 \( r \rightarrow 1 \) 时,分子和分母都趋向于 0,因此我们再次得到不定式 \( \frac{0}{0} \)。为了进一步分析,我们可以考虑分子和分母的比值: \[ \frac{r^2 \cos \theta \sin \theta - r \cos \theta - r \sin \theta + 1}{r^2 - 2 r \cos \theta - 2 r \sin \theta + 2}. \] ### 步骤 5:使用泰勒展开 我们可以对分子和分母进行泰勒展开。首先,泰勒展开 \( \cos \theta \) 和 \( \sin \theta \) 在 \( \theta = 0 \) 附近: \[ \cos \theta \approx 1 - \frac{\theta^2}{2}, \quad \sin \theta \approx \theta. \] 将这些展开式代入分子和分母中: \[ \cos \theta \sin \theta \approx (1 - \frac{\theta^2}{2}) \theta = \theta - \frac{\theta^3}{2}, \] \[ r^2 \cos \theta \sin \theta \approx r^2 (\theta - \frac{\theta^3}{2}) = r^2 \theta - \frac{r^2 \theta^3}{2}. \] 分子变为: \[ r^2 \theta - \frac{r^2 \theta^3}{2} - r \cos \theta - r \sin \theta + 1 \approx r^2 \theta - \frac{r^2 \theta^3}{2} - r (1 - \frac{\theta^2}{2}) - r \theta + 1. \] 分母变为: \[ r^2 - 2 r \cos \theta - 2 r \sin \theta + 2 \approx r^2 - 2 r (1 - \frac{\theta^2}{2}) - 2 r \theta + 2 = r^2 - 2 r + r \theta^2 - 2 r \theta + 2. \] ### 步骤 6:进一步简化 当 \( r \rightarrow 1 \) 时,我们可以忽略高阶项: \[ \operatorname { Lim } _ { r \rightarrow 1 } \frac { r^2 \theta - \frac{r^2 \theta

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\[ \operatorname { Lim } _ { ( x , y ) \rightarrow ( 1,1 ) } \frac { x y - x - y + 1 } { x ^ { 2 } + y ^ { 2 } - 2 x - 2 y + 2 } = \frac{1}{2} \]
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