| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The limit \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\) can be approached by considering what happens as \(x\) approaches 0 from both the positive and negative sides. Let's start by considering when \(x\) is positive and approaches 0. The floor function \(\lfloor 1/x \rfloor\) represents the greatest integer less than or equal to \(1/x\). As \(x\) approaches 0 from the positive side, \(1/x\) grows without bound, and \(\lfloor 1/x \rfloor\) will be a very large positive integer. However, since \(x\) is approaching 0, multiplying this large integer by \(x\) will result in a value that approaches 1. This is because the product will be a number just below 1 (since \(1/x\) will be just below the next integer, and multiplying it by \(x\) will make it just below 1). Now, let's consider when \(x\) is negative and approaches 0. In this case, \(1/x\) is a very large negative number. The floor function \(\lfloor 1/x \rfloor\) will be the greatest integer less than or equal to \(1/x\), which will be a very large negative integer. As \(x\) approaches 0 from the negative side, \(1/x\) approaches negative infinity, and \(\lfloor 1/x \rfloor\) will also approach negative infinity. However, when multiplied by \(x\) (which is negative), the product will approach 1 from the positive side. Thus, whether \(x\) approaches 0 from the positive or negative side, the limit of \(x \lfloor 1/x \rfloor\) will be 1. Hence, the solution is: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 1 \] Note: This result assumes that we are considering the limit in the real number system. In the complex plane, the limit may not exist due to the discontinuous nature of the floor function. However, in the context of real analysis, the limit is indeed 1. |
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| deepseekr1 | To find the limit \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\), we consider both the right-hand and left-hand limits as \(x\) approaches 0. ### Right-Hand Limit (x approaches 0 from the positive side) 1. As \(x \to 0^+\), \(1/x \to +\infty\). 2. Let \(n = \lfloor 1/x \rfloor\). Then, \(n \leq 1/x < n + 1\), which implies \(1/(n + 1) < x \leq 1/n\). 3. Multiplying by \(x\), we get \(x \cdot n \leq x \cdot (1/x) = 1\) and \(x \cdot n > x \cdot (1/x - 1) = 1 - x\). 4. As \(x \to 0^+\), \(n \to \infty\). Therefore, \(x \cdot n\) is squeezed between \(n/(n + 1)\) and 1. By the squeeze theorem, \(\lim_{x \to 0^+} x \lfloor 1/x \rfloor = 1\). ### Left-Hand Limit (x approaches 0 from the negative side) 1. As \(x \to 0^-\), \(1/x \to -\infty\). 2. Let \(n = \lfloor 1/x \rfloor\). Then, \(n \leq 1/x < n + 1\), where \(n\) is a negative integer. 3. Let \(n = -k\) where \(k\) is a positive integer. Then, \(-k \leq 1/x < -k + 1\), which implies \(-1/(k - 1) < x \leq -1/k\). 4. Multiplying by \(x\), we get \(x \cdot (-k) \geq 1\) and \(x \cdot (-k) < k/(k - 1)\). 5. As \(x \to 0^-\), \(k \to \infty\). Therefore, \(x \cdot (-k)\) is squeezed between 1 and \(k/(k - 1)\). By the squeeze theorem, \(\lim_{x \to 0^-} x \lfloor 1/x \rfloor = 1\). ### Conclusion Both the right-hand and left-hand limits are 1. Therefore, the overall limit is: \[ \boxed{1} \] |
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| gpto1 | To find the limit: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) \] we need to evaluate the behavior of the function \( x \left\lfloor \frac{1}{x} \right\rfloor \) as \( x \) approaches zero. Here, \( \left\lfloor \cdot \right\rfloor \) denotes the floor function, which returns the greatest integer less than or equal to its argument. ### Understanding the Behavior of the Function First, let's consider \( x \) approaching zero from the positive side (\( x \to 0^+ \)) and from the negative side (\( x \to 0^- \)) separately. #### Case 1: \( x \to 0^+ \) When \( x \) is a small positive number, \( \frac{1}{x} \) is a large positive number. Therefore, the floor function \( \left\lfloor \frac{1}{x} \right\rfloor \) gives a large positive integer. Let \( \frac{1}{x} = n + \epsilon \), where: - \( n = \left\lfloor \frac{1}{x} \right\rfloor \) is an integer. - \( \epsilon = \frac{1}{x} - n \) satisfies \( 0 \leq \epsilon < 1 \). Then: \[ x \left\lfloor \frac{1}{x} \right\rfloor = x n = x \left( \frac{1}{x} - \epsilon \right) = 1 - x \epsilon \] As \( x \to 0^+ \), \( x \epsilon \to 0 \) because \( 0 \leq \epsilon < 1 \) and \( x \to 0 \). Therefore: \[ \lim_{x \to 0^+} x \left\lfloor \frac{1}{x} \right\rfloor = 1 - 0 = 1 \] #### Case 2: \( x \to 0^- \) When \( x \) is a small negative number, \( \frac{1}{x} \) is a large negative number. The floor function \( \left\lfloor \frac{1}{x} \right\rfloor \) gives a large negative integer. Let \( \frac{1}{x} = -n + \epsilon \), where: - \( n = \left\lfloor \frac{1}{x} \right\rfloor \) is an integer (note that \( n \) is negative in this case). - \( \epsilon = -\frac{1}{x} - n \) satisfies \( 0 \leq \epsilon < 1 \). Then: \[ x \left\lfloor \frac{1}{x} \right\rfloor = x (-n) = -x n = -x \left( -\frac{1}{x} - \epsilon \right) = 1 - x \epsilon \] Again, as \( x \to 0^- \), \( x \epsilon \to 0 \) (since \( x \) is negative and \( \epsilon \geq 0 \), \( x \epsilon \to 0 \) from the positive side). Therefore: \[ \lim_{x \to 0^-} x \left\lfloor \frac{1}{x} \right\rfloor = 1 - 0 = 1 \] ### Conclusion Since both one-sided limits agree and equal 1, the two-sided limit is: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) = 1 \] **Answer:** The limit is **1**. |
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| GPT-4o | To find the limit \(\lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right)\), let's analyze the expression inside the limit. First, note that the floor function \(\left\lfloor \frac{1}{x} \right\rfloor\) denotes the greatest integer less than or equal to \(\frac{1}{x}\). Consider \(\frac{1}{x}\) as \(x\) approaches 0 from the positive side (\(x \to 0^+\)): - When \(x\) is a small positive number, \(\frac{1}{x}\) is a large positive number. - The floor function, \(\left\lfloor \frac{1}{x} \right\rfloor\), will be an integer very close to \(\frac{1}{x}\). More precisely, for \(x > 0\): \[ \left\lfloor \frac{1}{x} \right\rfloor \leq \frac{1}{x} < \left\lfloor \frac{1}{x} \right\rfloor + 1 \] Multiplying through by \(x\): \[ x \left\lfloor \frac{1}{x} \right\rfloor \leq x \cdot \frac{1}{x} < x \left( \left\lfloor \frac{1}{x} \right\rfloor + 1 \right) \] This simplifies to: \[ x \left\lfloor \frac{1}{x} \right\rfloor \leq 1 < x \left\lfloor \frac{1}{x} \right\rfloor + x \] Since \(x\) is approaching 0, the term \(x\) becomes very small. Therefore, the inequality \(1 < x \left\lfloor \frac{1}{x} \right\rfloor + x\) implies that as \(x \to 0\), the additional \(x\) term on the right-hand side also approaches 0. Hence, the expression \(x \left\lfloor \frac{1}{x} \right\rfloor\) gets squeezed between 1 and something slightly greater than 1, which approaches 1 as \(x\) approaches 0. Thus, we conclude: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) = 1 \] So, the solution to the given problem is: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) = 1 \] |
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| llama3.1-70B-Instruct | A nice limit problem! Let's analyze the expression: $$x \lfloor 1/x \rfloor$$ As $x$ approaches $0$, the value of $1/x$ increases without bound. Therefore, the floor function $\lfloor 1/x \rfloor$ will take on larger and larger integer values. Now, let's consider the product $x \lfloor 1/x \rfloor$. As $x$ approaches $0$, the factor $x$ approaches $0$, while the factor $\lfloor 1/x \rfloor$ approaches $\infty$. However, the product of a number approaching $0$ and a number approaching $\infty$ is not necessarily $0$ or $\infty$. We need to be more careful. Notice that $\lfloor 1/x \rfloor$ is always an integer, so we can write: $$x \lfloor 1/x \rfloor = x \cdot n$$ where $n$ is an integer. As $x$ approaches $0$, the value of $n$ increases without bound, but the product $x \cdot n$ approaches $0$ because $x$ is approaching $0$. Therefore, we can conclude that: $$\lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0$$ So, the limit is $0$. |
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| llama3.1-405B-Instruct | To evaluate the limit \lim_{x \to 0} (x \lfloor 1/x \rfloor), we need to consider the behavior of the function as x approaches 0 from both the left and the right. First, let's consider the case where x approaches 0 from the right, i.e., x > 0. In this case, we have: \lfloor 1/x \rfloor = \lfloor 1/(x+0) \rfloor = \lfloor 1/x \rfloor Since x is positive, we have: x \lfloor 1/x \rfloor = x \cdot \lfloor 1/x \rfloor Now, as x approaches 0 from the right, we have: \lim_{x \to 0^+} (x \lfloor 1/x \rfloor) = \lim_{x \to 0^+} (x \cdot \lfloor 1/x \rfloor) Since \lfloor 1/x \rfloor is an integer, we can write: \lfloor 1/x \rfloor = n where n is an integer. Then, we have: \lim_{x \to 0^+} (x \lfloor 1/x \rfloor) = \lim_{x \to 0^+} (x \cdot n) As x approaches 0 from the right, we have: \lim_{x \to 0^+} (x \cdot n) = 0 \cdot n = 0 Therefore, we have: \lim_{x \to 0^+} (x \lfloor 1/x \rfloor) = 0 Now, let's consider the case where x approaches 0 from the left, i.e., x < 0. In this case, we have: \lfloor 1/x \rfloor = \lfloor 1/(x-0) \rfloor = \lfloor 1/x \rfloor Since x is negative, we have: x \lfloor 1/x \rfloor = -x \cdot \lfloor 1/x \rfloor Now, as x approaches 0 from the left, we have: \lim_{x \to 0^-} (x \lfloor 1/x \rfloor) = \lim_{x \to 0^-} (-x \cdot \lfloor 1/x \rfloor) Since \lfloor 1/x \rfloor is an integer, we can write: \lfloor 1/x \rfloor = n where n is an integer. Then, we have: \lim_{x \to 0^-} (x \lfloor 1/x \rfloor) = \lim_{x \to 0^-} (-x \cdot n) As x approaches 0 from the left, we have: \lim_{x \to 0^-} (-x \cdot n) = 0 \cdot n = 0 Therefore, we have: \lim_{x \to 0^-} (x \lfloor 1/x \rfloor) = 0 Since the limits from both the left and the right are equal to 0, we can conclude that: \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0 Therefore, the final answer is 0. |
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| DeepSeek-Coder-V2 | To solve the problem \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\), we need to analyze the behavior of the expression as \(x\) approaches 0. 1. **Understanding the Floor Function**: The floor function \(\lfloor y \rfloor\) returns the greatest integer less than or equal to \(y\). For any positive \(y\), \(\lfloor y \rfloor\) is the integer part of \(y\). 2. **Expression Analysis**: Consider the expression \(x \lfloor 1/x \rfloor\). As \(x\) approaches 0 from the right (i.e., \(x > 0\)), \(1/x\) becomes very large. Specifically, for any small positive \(x\), \(1/x\) is a large positive number. 3. **Behavior of \(\lfloor 1/x \rfloor\)**: As \(x \to 0^+\), \(1/x \to \infty\). The floor function \(\lfloor 1/x \rfloor\) will be the greatest integer less than or equal to \(1/x\). For any small \(x\), \(\lfloor 1/x \rfloor\) will be a very large positive integer. 4. **Multiplying by \(x\)**: Now, consider the product \(x \lfloor 1/x \rfloor\). Since \(\lfloor 1/x \rfloor\) is a large positive integer, multiplying it by the small positive number \(x\) will result in a product that is still very close to 0. 5. **Limit Calculation**: Therefore, as \(x \to 0^+\), the product \(x \lfloor 1/x \rfloor\) approaches 0. Similarly, as \(x \to 0^-\) (i.e., \(x\) approaches 0 from the left), \(1/x\) becomes a very large negative number, and \(\lfloor 1/x \rfloor\) will be a very large negative integer. Multiplying this by the small negative number \(x\) will also result in a product that approaches 0. 6. **Conclusion**: Since the limit from both the right and the left sides approaches 0, we conclude that: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0 \] Thus, the solution to the problem is \(\boxed{0}\). |
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| Qwen2-72B-Instruct | The limit \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\) can be approached by considering what happens as \(x\) approaches 0 from both the positive and negative sides. Let's start by considering when \(x\) is positive and approaches 0. The floor function \(\lfloor 1/x \rfloor\) represents the greatest integer less than or equal to \(1/x\). As \(x\) approaches 0 from the positive side, \(1/x\) grows without bound, and \(\lfloor 1/x \rfloor\) will be a very large positive integer. However, since \(x\) is approaching 0, multiplying this large integer by \(x\) will result in a value that approaches 1. This is because the product will be a number just below 1 (since \(1/x\) will be just below the next integer, and multiplying it by \(x\) will make it just below 1). Now, let's consider when \(x\) is negative and approaches 0. In this case, \(1/x\) is a very large negative number. The floor function \(\lfloor 1/x \rfloor\) will be the greatest integer less than or equal to \(1/x\), which will be a very large negative integer. As \(x\) approaches 0 from the negative side, \(1/x\) approaches negative infinity, and \(\lfloor 1/x \rfloor\) will also approach negative infinity. However, when multiplied by \(x\) (which is negative), the product will approach 1 from the positive side. Thus, whether \(x\) approaches 0 from the positive or negative side, the limit of \(x \lfloor 1/x \rfloor\) will be 1. Hence, the solution is: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 1 \] Note: This result assumes that we are considering the limit in the real number system. In the complex plane, the limit may not exist due to the discontinuous nature of the floor function. However, in the context of real analysis, the limit is indeed 1. |
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| baichuan4 | To find the limit of $x \lfloor 1/x \rfloor$ as $x$ approaches $0$, we need to consider the behavior of the expression as $x$ gets very small. First, let's analyze the behavior of $\lfloor 1/x \rfloor$ for different values of $x$. When $x$ is a positive number close to $0$, $1/x$ becomes a large positive number. Since $\lfloor 1/x \rfloor$ is the floor function applied to $1/x$, it will give us the largest integer less than or equal to $1/x$. Now, let's consider two cases: 1. When $0 < x < 1$, then $1/x > 1$, and $\lfloor 1/x \rfloor$ will be $1$ or a larger positive integer. 2. When $-1 < x < 0$, then $1/x$ is negative, and $\lfloor 1/x \rfloor$ will be $-1$ or a more negative integer. Now, let's multiply these values by $x$: 1. If $0 < x < 1$, then $x \lfloor 1/x \rfloor$ will be $x$ times a positive integer, which will be positive and approach $0$ as $x$ approaches $0$ from the right. 2. If $-1 < x < 0$, then $x \lfloor 1/x \rfloor$ will be $x$ times a negative integer, which will be negative and approach $0$ as $x$ approaches $0$ from the left. Since the left-hand limit and the right-hand limit both approach $0$, we can conclude that: $$\lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0$$ |
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| doubao | As x approaches 0, 1/x approaches infinity. The floor function ⌊1/x⌋ will take on increasingly large integer values. Now consider x⌊1/x⌋. As x approaches 0, the product will also approach 0 because the multiplying factor x is approaching 0 while the ⌊1/x⌋ term is bounded. So, the limit as x approaches 0 of x⌊1/x⌋ is 0. |
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| minimax | To solve the limit \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\), we need to understand the behavior of the floor function \(\lfloor 1/x \rfloor\) as \(x\) approaches 0. The floor function \(\lfloor y \rfloor\) gives the greatest integer less than or equal to \(y\). As \(x\) approaches 0 from the right (\(x \to 0^+\)), \(1/x\) becomes very large, and the floor of \(1/x\) will be a sequence of integers that increase without bound. Similarly, as \(x\) approaches 0 from the left (\(x \to 0^-\)), \(1/x\) becomes very large in the negative direction, and the floor of \(1/x\) will be a sequence of negative integers that decrease without bound. However, the product \(x \lfloor 1/x \rfloor\) will always be an integer because \(x\) is a real number and \(\lfloor 1/x \rfloor\) is an integer. As \(x\) gets closer and closer to 0, the magnitude of the integer part \(\lfloor 1/x \rfloor\) increases, but since \(x\) is also getting smaller, the product \(x\) times this large integer will approach 0. To be more precise, for any \(x\) close to 0, the product \(x \lfloor 1/x \rfloor\) will be less than or equal to \(x\) in absolute value because the absolute value of the floor function is always less than or equal to the absolute value of the number itself. Therefore, as \(x\) approaches 0, the absolute value of the product \(x \lfloor 1/x \rfloor\) will be less than or equal to the absolute value of \(x\), which approaches 0. Thus, the limit is: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0 \] This is because the product \(x \lfloor 1/x \rfloor\) will be squeezed between \(-|x|\) and \(|x|\), and as \(x\) approaches 0, both of these bounds approach 0. Therefore, by the squeeze theorem, the limit of \(x \lfloor 1/x \rfloor\) as \(x\) approaches 0 is 0. |
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| kimi | To solve the limit problem \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\), we need to consider the behavior of the function as \(x\) approaches 0 from both the positive and negative sides. The floor function \(\lfloor y \rfloor\) returns the greatest integer less than or equal to \(y\). As \(x\) approaches 0 from the positive side (\(x \to 0^+\)), \(1/x\) approaches infinity, and the floor function \(\lfloor 1/x \rfloor\) will be a large positive integer \(n\). Therefore, \(x \lfloor 1/x \rfloor\) will be a small positive number \(n \cdot x\), which approaches 0 as \(x\) approaches 0. Similarly, as \(x\) approaches 0 from the negative side (\(x \to 0^-\)), \(1/x\) approaches negative infinity, and the floor function \(\lfloor 1/x \rfloor\) will be a large negative integer \(-n\). Therefore, \(x \lfloor 1/x \rfloor\) will be a small negative number \(-n \cdot x\), which also approaches 0 as \(x\) approaches 0. Since the limit from both sides is the same, we can conclude that: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 0 \] |
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| wenxinyiyan | 为了解决这个问题，我们需要考虑函数$f(x) = x \lfloor 1/x \rfloor$在$x$接近$0$时的行为。 首先，我们观察函数$\lfloor 1/x \rfloor$。当$x$接近$0$时，$1/x$的绝对值会趋向于无穷大。因此，$\lfloor 1/x \rfloor$会趋向于无穷大或者无穷小，取决于$x$是从正方向还是负方向接近$0$。 现在，我们考虑$x \lfloor 1/x \rfloor$。当$x$从正方向接近$0$时，$x$是正的，而$\lfloor 1/x \rfloor$趋向于正无穷大，所以它们的乘积也趋向于正无穷大。当$x$从负方向接近$0$时，$x$是负的，而$\lfloor 1/x \rfloor$趋向于负无穷大，所以它们的乘积同样趋向于正无穷大。 因此，无论是从正方向还是负方向接近$0$，函数$f(x) = x \lfloor 1/x \rfloor$都趋向于正无穷大。所以，极限$\lim_{x \to 0} (x \lfloor 1/x \rfloor)$不存在，因为函数没有收敛到一个有限的值。 综上所述，$\lim_{x \to 0} (x \lfloor 1/x \rfloor)$不存在。 |
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| zhipu | To solve the limit \(\lim_{x \to 0} (x \lfloor 1/x \rfloor)\), we need to understand the behavior of the function \(x \lfloor 1/x \rfloor\) as \(x\) approaches 0. First, let's break down the components: - \(x\) is the variable approaching 0. - \(\lfloor 1/x \rfloor\) is the floor function applied to \(1/x\). The floor function \(\lfloor y \rfloor\) returns the greatest integer less than or equal to \(y\). So, \(\lfloor 1/x \rfloor\) will be the greatest integer less than or equal to \(1/x\). ### Analysis for \(x > 0\): When \(x\) is a small positive number: - \(1/x\) becomes very large. - \(\lfloor 1/x \rfloor\) will be the largest integer less than or equal to \(1/x\), which is approximately \(1/x\) for very small \(x\). Thus, for \(x > 0\): \[ x \lfloor 1/x \rfloor \approx x \cdot \frac{1}{x} = 1. \] ### Analysis for \(x < 0\): When \(x\) is a small negative number: - \(1/x\) becomes a very large negative number. - \(\lfloor 1/x \rfloor\) will be the largest integer less than or equal to \(1/x\), which is approximately \(1/x\) for very small \(x\). Thus, for \(x < 0\): \[ x \lfloor 1/x \rfloor \approx x \cdot \frac{1}{x} = 1. \] ### Combining Both Cases: As \(x\) approaches 0 from both the positive and negative sides, \(x \lfloor 1/x \rfloor\) approaches 1. Therefore, the limit is: \[ \lim_{x \to 0} (x \lfloor 1/x \rfloor) = 1. \] ### Conclusion: \[ \boxed{1} \] |
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| gpto1 | To find the limit: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) \] we need to evaluate the behavior of the function \( x \left\lfloor \frac{1}{x} \right\rfloor \) as \( x \) approaches zero. Here, \( \left\lfloor \cdot \right\rfloor \) denotes the floor function, which returns the greatest integer less than or equal to its argument. ### Understanding the Behavior of the Function First, let's consider \( x \) approaching zero from the positive side (\( x \to 0^+ \)) and from the negative side (\( x \to 0^- \)) separately. #### Case 1: \( x \to 0^+ \) When \( x \) is a small positive number, \( \frac{1}{x} \) is a large positive number. Therefore, the floor function \( \left\lfloor \frac{1}{x} \right\rfloor \) gives a large positive integer. Let \( \frac{1}{x} = n + \epsilon \), where: - \( n = \left\lfloor \frac{1}{x} \right\rfloor \) is an integer. - \( \epsilon = \frac{1}{x} - n \) satisfies \( 0 \leq \epsilon < 1 \). Then: \[ x \left\lfloor \frac{1}{x} \right\rfloor = x n = x \left( \frac{1}{x} - \epsilon \right) = 1 - x \epsilon \] As \( x \to 0^+ \), \( x \epsilon \to 0 \) because \( 0 \leq \epsilon < 1 \) and \( x \to 0 \). Therefore: \[ \lim_{x \to 0^+} x \left\lfloor \frac{1}{x} \right\rfloor = 1 - 0 = 1 \] #### Case 2: \( x \to 0^- \) When \( x \) is a small negative number, \( \frac{1}{x} \) is a large negative number. The floor function \( \left\lfloor \frac{1}{x} \right\rfloor \) gives a large negative integer. Let \( \frac{1}{x} = -n + \epsilon \), where: - \( n = \left\lfloor \frac{1}{x} \right\rfloor \) is an integer (note that \( n \) is negative in this case). - \( \epsilon = -\frac{1}{x} - n \) satisfies \( 0 \leq \epsilon < 1 \). Then: \[ x \left\lfloor \frac{1}{x} \right\rfloor = x (-n) = -x n = -x \left( -\frac{1}{x} - \epsilon \right) = 1 - x \epsilon \] Again, as \( x \to 0^- \), \( x \epsilon \to 0 \) (since \( x \) is negative and \( \epsilon \geq 0 \), \( x \epsilon \to 0 \) from the positive side). Therefore: \[ \lim_{x \to 0^-} x \left\lfloor \frac{1}{x} \right\rfloor = 1 - 0 = 1 \] ### Conclusion Since both one-sided limits agree and equal 1, the two-sided limit is: \[ \lim_{x \to 0} \left( x \left\lfloor \frac{1}{x} \right\rfloor \right) = 1 \] **Answer:** The limit is **1**. |
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