| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the limit \(\lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1}))\), we can use the Squeeze Theorem. The Squeeze Theorem (also known as the Sandwich Theorem) states that if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in an open interval containing \(a\), except possibly at \(a\) itself, and if \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} g(x) = L\). Let's consider the function \(g(x) = (x - 1) \sin(\frac{\pi}{x - 1})\). We need to find two functions \(f(x)\) and \(h(x)\) such that \(f(x) \leq g(x) \leq h(x)\) and \(\lim_{x \to 1} f(x) = \lim_{x \to 1} h(x)\). We know that \(-1 \leq \sin(\theta) \leq 1\) for any real number \(\theta\). Therefore, we have: \[ -1 \leq \sin\left(\frac{\pi}{x - 1}\right) \leq 1 \] Multiplying the inequality by \((x - 1)\), we get: \[ -(x - 1) \leq (x - 1) \sin\left(\frac{\pi}{x - 1}\right) \leq (x - 1) \] Let's define \(f(x) = -(x - 1)\) and \(h(x) = (x - 1)\). We can see that: \[ f(x) \leq g(x) \leq h(x) \] Now, we need to find the limits of \(f(x)\) and \(h(x)\) as \(x\) approaches 1. We have: \[ \lim_{x \to 1} f(x) = \lim_{x \to 1} -(x - 1) = 0 \] \[ \lim_{x \to 1} h(x) = \lim_{x \to 1} (x - 1) = 0 \] Since \(\lim_{x \to 1} f(x) = \lim_{x \to 1} h(x) = 0\), by the Squeeze Theorem, we have: \[ \lim_{x \to 1} g(x) = \lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right)) = 0 \] Therefore, the solution to the given problem is \(\boxed{0}\). |
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| deepseekr1 | To find the limit \(\lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right)\), we start by substituting \(t = x - 1\). As \(x\) approaches 1, \(t\) approaches 0. This transforms the limit into: \[ \lim_{t \to 0} t \sin\left(\frac{\pi}{t}\right) \] Next, we analyze the behavior of the function. The sine function \(\sin\left(\frac{\pi}{t}\right)\) oscillates between -1 and 1 as \(t\) approaches 0. The term \(t\) approaches 0. The product of a term approaching 0 and a bounded function (between -1 and 1) will approach 0. To apply the squeeze theorem, note that for all \(t \neq 0\): \[ -1 \leq \sin\left(\frac{\pi}{t}\right) \leq 1 \] Multiplying all parts by \(t\) (considering both positive and negative values of \(t\)): - If \(t > 0\), then \(-t \leq t \sin\left(\frac{\pi}{t}\right) \leq t\) - If \(t < 0\), then \(t \leq t \sin\left(\frac{\pi}{t}\right) \leq -t\) In both cases, as \(t \to 0\), both \(-|t|\) and \(|t|\) approach 0. By the squeeze theorem, the expression \(t \sin\left(\frac{\pi}{t}\right)\) is squeezed between these values and must also approach 0. Thus, the limit is: \[ \boxed{0} \] |
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| gpto1 | To evaluate the limit \[ L = \lim_{x \to 1} (x - 1) \sin\left(\frac{\pi}{x - 1}\right), \] we can make a substitution to simplify the expression. Let \[ h = x - 1, \] so when \( x \to 1 \), \( h \to 0 \). The limit becomes \[ L = \lim_{h \to 0} h \sin\left(\frac{\pi}{h}\right). \] We observe that as \( h \to 0 \), \( \frac{\pi}{h} \to \infty \), and the sine function oscillates between \(-1\) and \(1\). Specifically, \[ -1 \le \sin\left(\frac{\pi}{h}\right) \le 1. \] Multiplying all parts of the inequality by \( h \) (which is approaching zero), we obtain \[ - h \le h \sin\left(\frac{\pi}{h}\right) \le h. \] As \( h \to 0 \), both \( -h \) and \( h \) approach zero. Therefore, by the Squeeze Theorem, the expression \( h \sin\left(\frac{\pi}{h}\right) \) must also approach zero. Formally, \[ \lim_{h \to 0} (- h) = 0, \quad \lim_{h \to 0} h = 0, \] so \[ \lim_{h \to 0} h \sin\left(\frac{\pi}{h}\right) = 0. \] **Answer:** 0 |
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| GPT-4o | To determine the limit \(\lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right)\), we need to analyze the behavior of the function as \(x\) approaches 1. 1. **Substitution and Simplification:** Let \( t = x - 1 \). As \( x \to 1 \), \( t \to 0 \). The limit then transforms as follows: \[ \lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right) = \lim_{t \to 0} \left( t \sin\left(\frac{\pi}{t}\right)\right). \] 2. **Analysis of the Expression:** Let's consider the expression \( t \sin\left(\frac{\pi}{t}\right) \). We know the following properties of the sine function: - The sine function is bounded, i.e., \( -1 \leq \sin\left(\frac{\pi}{t}\right) \leq 1 \). 3. **Bounding the Expression:** Since \( \sin\left(\frac{\pi}{t}\right) \) is bounded, we can write: \[ -|t| \leq t \sin\left(\frac{\pi}{t}\right) \leq |t|. \] 4. **Applying the Squeeze Theorem:** As \( t \to 0 \), both \(-|t|\) and \(|t|\) approach 0. Therefore, by the Squeeze Theorem: \[ \lim_{t \to 0} t \sin\left(\frac{\pi}{t}\right) = 0. \] Thus, the limit is: \[ \lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right) = 0. \] So, the solution is: \[ \boxed{0}. \] |
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| llama3.1-70B-Instruct | A nice limit problem! Let's analyze the given expression: $$\lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right))$$ As $x$ approaches $1$, the term $(x - 1)$ approaches $0$. However, the term $\sin\left(\frac{\pi}{x - 1}\right)$ approaches $\sin(\infty)$, which is undefined. To tackle this, we can use the fact that $\sin(u)$ is bounded between $-1$ and $1$ for any real number $u$. In particular, we have: $$-1 \leq \sin\left(\frac{\pi}{x - 1}\right) \leq 1$$ Now, multiply all three parts of the inequality by $(x - 1)$, which is non-negative since $x$ is approaching $1$ from the right: $$-(x - 1) \leq (x - 1) \sin\left(\frac{\pi}{x - 1}\right) \leq (x - 1)$$ As $x$ approaches $1$, the left-hand side and right-hand side of the inequality both approach $0$. By the Squeeze Theorem, we conclude that: $$\lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right)) = 0$$ So, the limit is $0$. |
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| llama3.1-405B-Instruct | To evaluate the limit \lim_{x \to 1} ((x - 1) sin((\pi)/(x - 1)), we can use the following steps: 1. Factor out (x - 1) from the expression: \lim_{x \to 1} ((x - 1) sin((\pi)/(x - 1)) = \lim_{x \to 1} (x - 1) \cdot \lim_{x \to 1} sin((\pi)/(x - 1)) 2. Evaluate the limit of (x - 1) as x approaches 1: \lim_{x \to 1} (x - 1) = 1 - 1 = 0 3. Evaluate the limit of sin((\pi)/(x - 1)) as x approaches 1: As x approaches 1, the expression (\pi)/(x - 1) approaches infinity. Since the sine function oscillates between -1 and 1 as its argument approaches infinity, we can conclude that: \lim_{x \to 1} sin((\pi)/(x - 1)) = 0 4. Combine the results from steps 2 and 3: \lim_{x \to 1} ((x - 1) sin((\pi)/(x - 1)) = 0 \cdot 0 = 0 Therefore, the limit is 0. |
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| DeepSeek-Coder-V2 | To solve the limit \(\lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right))\), we can use the Squeeze Theorem. Here is the step-by-step solution: 1. **Identify the function and the limit:** \[ \lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right)) \] 2. **Understand the behavior of the sine function:** The sine function is bounded between -1 and 1 for all real numbers: \[ -1 \leq \sin\left(\frac{\pi}{x - 1}\right) \leq 1 \] 3. **Multiply the inequality by \((x - 1)\):** Since \(x \to 1\), \(x - 1\) approaches 0. Therefore, we can multiply the inequality by \((x - 1)\): \[ -(x - 1) \leq (x - 1) \sin\left(\frac{\pi}{x - 1}\right) \leq (x - 1) \] 4. **Evaluate the limits of the bounding functions:** As \(x \to 1\), \((x - 1) \to 0\). Thus: \[ \lim_{x \to 1} -(x - 1) = 0 \] \[ \lim_{x \to 1} (x - 1) = 0 \] 5. **Apply the Squeeze Theorem:** Since \((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\) is squeezed between \(-(x - 1)\) and \((x - 1)\), and both of these limits as \(x \to 1\) are 0, by the Squeeze Theorem: \[ \lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right)) = 0 \] Therefore, the solution to the limit is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To solve the limit \(\lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1}))\), we can use the Squeeze Theorem. The Squeeze Theorem (also known as the Sandwich Theorem) states that if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in an open interval containing \(a\), except possibly at \(a\) itself, and if \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} g(x) = L\). Let's consider the function \(g(x) = (x - 1) \sin(\frac{\pi}{x - 1})\). We need to find two functions \(f(x)\) and \(h(x)\) such that \(f(x) \leq g(x) \leq h(x)\) and \(\lim_{x \to 1} f(x) = \lim_{x \to 1} h(x)\). We know that \(-1 \leq \sin(\theta) \leq 1\) for any real number \(\theta\). Therefore, we have: \[ -1 \leq \sin\left(\frac{\pi}{x - 1}\right) \leq 1 \] Multiplying the inequality by \((x - 1)\), we get: \[ -(x - 1) \leq (x - 1) \sin\left(\frac{\pi}{x - 1}\right) \leq (x - 1) \] Let's define \(f(x) = -(x - 1)\) and \(h(x) = (x - 1)\). We can see that: \[ f(x) \leq g(x) \leq h(x) \] Now, we need to find the limits of \(f(x)\) and \(h(x)\) as \(x\) approaches 1. We have: \[ \lim_{x \to 1} f(x) = \lim_{x \to 1} -(x - 1) = 0 \] \[ \lim_{x \to 1} h(x) = \lim_{x \to 1} (x - 1) = 0 \] Since \(\lim_{x \to 1} f(x) = \lim_{x \to 1} h(x) = 0\), by the Squeeze Theorem, we have: \[ \lim_{x \to 1} g(x) = \lim_{x \to 1} ((x - 1) \sin\left(\frac{\pi}{x - 1}\right)) = 0 \] Therefore, the solution to the given problem is \(\boxed{0}\). |
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| baichuan4 | To find the limit as x approaches 1 of ((x - 1) sin((π)/(x - 1)), we can use L'Hôpital's Rule, since the function appears to be of the form "0 times something" when x is close to 1. Let's set the function we're taking the limit of as f(x): f(x) = (x - 1)sin((π)/(x - 1)) As x approaches 1, f(x) becomes: f(1) = (1 - 1)sin((π)/(1 - 1)) f(1) = 0sin(∞) This is an indeterminate form (0 * ∞), so we can apply L'Hôpital's Rule. We need to differentiate the numerator and denominator separately with respect to x: f'(x) = derivative of (x - 1)sin((π)/(x - 1)) Using the product rule, we get: f'(x) = sin((π)/(x - 1)) + (x - 1)cos((π)/(x - 1))*(-π/(x - 1)^2) Simplifying: f'(x) = sin((π)/(x - 1)) - πcos((π)/(x - 1))/(x - 1) Now, let's evaluate the limit as x approaches 1: lim (x→1) f'(x) = sin(π/0) - πcos(π/0)/0 This is still indeterminate (∞ - ∞), so we need to apply L'Hôpital's Rule again. We'll differentiate the numerator and denominator of f'(x) with respect to x: f''(x) = derivative of sin((π)/(x - 1)) - πcos((π)/(x - 1))/(x - 1) Using the chain rule on the first term and the quotient rule on the second, we get: f''(x) = -πcos((π)/(x - 1))/(x - 1)^2 - [(-πsin((π)/(x - 1))/(x - 1)^2)(-πcos((π)/(x - 1))/(x - 1)) - (πcos((π)/(x - 1))/(x - 1))(-πsin((π)/(x - 1))/(x - 1)^2)] / (x - 1)^2 Simplifying: f''(x) = -πcos((π)/(x - 1))/(x - 1)^2 - π^2cos((π)/(x - 1))sin((π)/(x - 1))/(x - 1)^4 + π^2sin((π)/(x - 1))cos((π)/(x - 1))/(x - 1)^4 Canceling out terms and evaluating at x = 1: lim (x→1) f''(x) = -πcos(∞)/0^2 This is still indeterminate (0/0), so we need to apply L'Hôpital's Rule once more. We'll differentiate the numerator and denominator of f''(x) with respect to x: f'''(x) = derivative of -πcos((π)/(x - 1))/(x - 1)^2 Using the chain rule and the power rule, we get: f'''(x) = πsin((π)/(x - 1))/(x - 1)^3 * (-π/(x - 1)^2) - 2(-πcos((π)/(x - 1))/(x - 1)^3) Simplifying: f'''(x) = -π^2sin((π)/(x - 1))/(x - 1)^5 + 2πcos((π)/(x - 1))/(x - 1)^3 Now, let's evaluate the limit as x approaches 1: lim (x→1) f'''(x) = -π^2sin(∞)/0^5 + 2πcos(∞)/0^3 This is still indeterminate (0/0), so we need to apply L'Hôpital's Rule again. We'll differentiate the numerator and denominator of f'''(x) with respect to x: f''''(x) = derivative of -π^2sin((π)/(x - 1))/(x - 1)^5 + 2πcos((π)/(x - 1))/(x - 1)^3 Using the chain rule, quotient rule, and power rule, we get: f''''(x) = -π^2cos((π)/(x - 1))/(x - 1)^6 * (-π/(x - 1)^2) - 5(-π^2sin((π)/(x - 1))/(x - 1)^6) + 2π(-sin((π)/(x - 1))/(x - 1)^4) - 3(2πcos((π)/(x - 1))/(x - 1)^4) Simplifying: f''''(x) = π^3cos((π)/(x - 1))/(x - 1)^8 + 5π^2sin((π)/(x - 1))/(x - 1)^6 - 2πsin((π)/(x - 1))/(x - 1)^4 - 6πcos((π)/(x - 1))/(x - 1)^4 Now, let's evaluate the limit as x approaches 1: lim (x→1) f''''(x) = π^3cos(∞)/0^8 + 5π^2sin(∞)/0^6 - 2πsin(∞)/0^4 - 6πcos(∞)/0^4 This is still indeterminate, but we can see a pattern emerging. Each time we apply L'Hôpital's Rule, the powers of (x - 1) in the denominator increase, making the terms closer to zero. If we continue this process indefinitely, all the terms will go to zero except for the first one, which will be: lim (x→1) π^3cos((π)/(x - 1))/(x - 1)^8 Since the cosine function is bounded between -1 and 1, this limit will also be zero: lim (x→1) π^3cos((π)/(x - 1))/(x - 1)^8 = 0 Therefore, the original limit is also zero: lim (x→1) ((x - 1)sin((π)/(x - 1)) = 0 |
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| doubao | As x approaches 1, (x - 1) approaches 0. And the sine function oscillates between -1 and 1. So the product (x - 1)sin(π/(x - 1)) will also approach 0 since the factor (x - 1) goes to 0 and the sine term is bounded. So, the limit is 0. |
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| minimax | To find the limit \(\lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1}))\), we can observe that as \(x\) approaches 1, the term \((x - 1)\) approaches 0, and the term \(\sin(\frac{\pi}{x - 1})\) oscillates between -1 and 1. However, since the sine function is bounded between -1 and 1, the product of \((x - 1)\) and \(\sin(\frac{\pi}{x - 1})\) will approach 0 as \(x\) approaches 1. This is because the factor \((x - 1)\) is the one that dominates the behavior of the product as \(x\) approaches 1. The sine function does not affect the limit because its absolute value is never greater than 1, and thus the product is squeezed between \(-|x - 1|\) and \(|x - 1|\), both of which approach 0 as \(x\) approaches 1. Therefore, the limit is: \[ \lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1})) = 0 \] |
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| kimi | To solve the limit \(\lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1}))\), we can use the Squeeze Theorem (also known as the Sandwich Theorem) or L'Hôpital's Rule. However, in this case, L'Hôpital's Rule is not directly applicable because the limit is of the form \(0/0\) or \(\infty/\infty\), which is not the case here. Instead, we can use the Squeeze Theorem. First, let's rewrite the expression in a more convenient form: \[ \lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1})) \] Notice that for all \(x\) in the domain of the function, the sine function is bounded between -1 and 1: \[ -1 \leq \sin(\frac{\pi}{x - 1}) \leq 1 \] Multiplying by \((x - 1)\), which approaches 0 as \(x\) approaches 1, we get: \[ -(x - 1) \leq (x - 1) \sin(\frac{\pi}{x - 1}) \leq (x - 1) \] As \(x\) approaches 1, the left-hand side and the right-hand side of the inequality both approach 0: \[ \lim_{x \to 1} -(x - 1) = 0 \quad \text{and} \quad \lim_{x \to 1} (x - 1) = 0 \] By the Squeeze Theorem, since the expression \((x - 1) \sin(\frac{\pi}{x - 1})\) is squeezed between two expressions that approach 0 as \(x\) approaches 1, the limit must be: \[ \lim_{x \to 1} ((x - 1) \sin(\frac{\pi}{x - 1})) = 0 \] So the solution to the limit is: \[ \boxed{0} \] |
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| wenxinyiyan | 首先，我们注意到当$x$接近$1$时，$\frac{\pi}{x - 1}$会变得非常大，这意味着$\sin\left(\frac{\pi}{x - 1}\right)$会快速振荡。但是，由于$x - 1$接近$0$，整个表达式$(x - 1) \sin\left(\frac{\pi}{x - 1}\right)$可能会收敛到一个确定的值。 为了解决这个问题，我们可以使用夹逼定理（Squeeze Theorem）。考虑以下不等式： $$0 \leq |(x - 1) \sin\left(\frac{\pi}{x - 1}\right)| \leq |x - 1|$$当$x$接近$1$时，$|x - 1|$接近$0$，因此根据夹逼定理，$\lim_{{x \to 1}} |(x - 1) \sin\left(\frac{\pi}{x - 1}\right)| = 0$。 由于绝对值函数的连续性，我们可以得出： $$\lim_{{x \to 1}} (x - 1) \sin\left(\frac{\pi}{x - 1}\right) = 0$$ |
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| zhipu | To find the limit \(\lim_{x \to 1} (x - 1) \sin\left(\frac{\pi}{x - 1}\right)\), we need to analyze the behavior of the function as \(x\) approaches 1. First, let's make a substitution to simplify the expression. Let \( t = x - 1 \). As \( x \to 1 \), \( t \to 0 \). The limit then becomes: \[ \lim_{t \to 0} t \sin\left(\frac{\pi}{t}\right). \] Now, let's examine the behavior of \( \sin\left(\frac{\pi}{t}\right) \) as \( t \to 0 \). The sine function oscillates between -1 and 1 for all real numbers, so: \[ -1 \leq \sin\left(\frac{\pi}{t}\right) \leq 1. \] Multiplying these inequalities by \( t \), we get: \[ -t \leq t \sin\left(\frac{\pi}{t}\right) \leq t. \] As \( t \to 0 \), both \(-t\) and \(t\) approach 0. By the Squeeze Theorem (also known as the Sandwich Theorem), if a function \( f(t) \) is squeezed between two functions that both approach the same limit, then \( f(t) \) must also approach that limit. Here, \( t \sin\left(\frac{\pi}{t}\right) \) is squeezed between \(-t\) and \(t\), both of which approach 0 as \( t \to 0 \). Therefore, by the Squeeze Theorem: \[ \lim_{t \to 0} t \sin\left(\frac{\pi}{t}\right) = 0. \] Substituting back \( t = x - 1 \), we have: \[ \lim_{x \to 1} (x - 1) \sin\left(\frac{\pi}{x - 1}\right) = 0. \] Thus, the solution to the problem is: \[ \boxed{0}. \] |
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| gpto1 | To evaluate the limit \[ L = \lim_{x \to 1} (x - 1) \sin\left(\frac{\pi}{x - 1}\right), \] we can make a substitution to simplify the expression. Let \[ h = x - 1, \] so when \( x \to 1 \), \( h \to 0 \). The limit becomes \[ L = \lim_{h \to 0} h \sin\left(\frac{\pi}{h}\right). \] We observe that as \( h \to 0 \), \( \frac{\pi}{h} \to \infty \), and the sine function oscillates between \(-1\) and \(1\). Specifically, \[ -1 \le \sin\left(\frac{\pi}{h}\right) \le 1. \] Multiplying all parts of the inequality by \( h \) (which is approaching zero), we obtain \[ - h \le h \sin\left(\frac{\pi}{h}\right) \le h. \] As \( h \to 0 \), both \( -h \) and \( h \) approach zero. Therefore, by the Squeeze Theorem, the expression \( h \sin\left(\frac{\pi}{h}\right) \) must also approach zero. Formally, \[ \lim_{h \to 0} (- h) = 0, \quad \lim_{h \to 0} h = 0, \] so \[ \lim_{h \to 0} h \sin\left(\frac{\pi}{h}\right) = 0. \] **Answer:** 0 |
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| deepseekv3 | To evaluate the limit \(\lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right)\), let's analyze the behavior of the expression as \(x\) approaches \(1\). ### Step 1: Substitute \(u = x - 1\) Let \(u = x - 1\). As \(x \to 1\), \(u \to 0\). The limit becomes: \[ \lim_{u \to 0} \left(u \sin\left(\frac{\pi}{u}\right)\right). \] ### Step 2: Analyze the behavior of \(\sin\left(\frac{\pi}{u}\right)\) As \(u \to 0\), \(\frac{\pi}{u}\) grows without bound, and \(\sin\left(\frac{\pi}{u}\right)\) oscillates between \(-1\) and \(1\) infinitely many times. ### Step 3: Multiply by \(u\) Since \(u \to 0\) and \(\sin\left(\frac{\pi}{u}\right)\) is bounded (between \(-1\) and \(1\)), the product \(u \sin\left(\frac{\pi}{u}\right)\) will also approach \(0\). This is because \(u\) dominates the oscillatory behavior of \(\sin\left(\frac{\pi}{u}\right)\). ### Step 4: Conclude the limit Thus, the limit is: \[ \lim_{x \to 1} \left((x - 1) \sin\left(\frac{\pi}{x - 1}\right)\right) = 0. \] ### Final Answer: \[ \boxed{0} \] |
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