| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the limit \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we can rewrite the expression using the definitions of \(\csc(x)\) and \(\cot(x)\): \[ \csc(x) = \frac{1}{\sin(x)} \] \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] So, \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = \lim_{x \to 0} \left( \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} \right) \] \[ = \lim_{x \to 0} \left( \frac{1 - \cos(x)}{\sin(x)} \right) \] To evaluate this limit, we can use L'Hôpital's rule since the limit takes the form \(\frac{0}{0}\) as \(x \to 0\). L'Hôpital's rule states that if the limit of the ratio of the derivatives of the numerator and denominator exists, then this limit is equal to the limit of the original ratio. Differentiating the numerator and the denominator with respect to \(x\), we get: Numerator: \(\frac{d}{dx} (1 - \cos(x)) = \sin(x)\) Denominator: \(\frac{d}{dx} \sin(x) = \cos(x)\) So, \[ \lim_{x \to 0} \left( \frac{1 - \cos(x)}{\sin(x)} \right) = \lim_{x \to 0} \left( \frac{\sin(x)}{\cos(x)} \right) \] \[ = \lim_{x \to 0} \tan(x) \] Since \(\tan(0) = 0\), \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = 0 \] Therefore, the solution to the given problem is \(\boxed{0}\). |
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| deepseekr1 | To find the limit as \( x \) approaches 0 of \( \csc(x) - \cot(x) \): 1. Rewrite the expression using trigonometric identities: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] 2. Simplify the expression by multiplying the numerator and denominator by \( 1 + \cos(x) \): \[ \frac{1 - \cos(x)}{\sin(x)} \cdot \frac{1 + \cos(x)}{1 + \cos(x)} = \frac{\sin^2(x)}{\sin(x)(1 + \cos(x))} = \frac{\sin(x)}{1 + \cos(x)} \] 3. Evaluate the limit as \( x \) approaches 0: \[ \lim_{x \to 0} \frac{\sin(x)}{1 + \cos(x)} = \frac{0}{1 + 1} = 0 \] 4. Alternatively, apply L'Hospital's Rule to the original indeterminate form \( \frac{1 - \cos(x)}{\sin(x)} \): \[ \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \lim_{x \to 0} \tan(x) = 0 \] 5. Using Taylor series expansions: \[ 1 - \cos(x) \approx \frac{x^2}{2} \quad \text{and} \quad \sin(x) \approx x \] \[ \frac{1 - \cos(x)}{\sin(x)} \approx \frac{\frac{x^2}{2}}{x} = \frac{x}{2} \to 0 \quad \text{as} \quad x \to 0 \] All methods confirm that the limit is \(\boxed{0}\). |
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| gpto1 | To find the limit: \[ \lim_{x \to 0} \left( \csc(x) - \cot(x) \right) \] First, recall the definitions of cosecant and cotangent in terms of sine and cosine: \[ \csc(x) = \frac{1}{\sin(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)} \] Subtracting the two gives: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] Now, consider the limit: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] We can simplify the numerator using the identity \(1 - \cos(x) = 2 \sin^2\left(\dfrac{x}{2}\right)\): \[ \frac{1 - \cos(x)}{\sin(x)} = \frac{2 \sin^2\left(\dfrac{x}{2}\right)}{2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right)} = \frac{\sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right)} = \tan\left(\dfrac{x}{2}\right) \] Therefore, the limit becomes: \[ \lim_{x \to 0} \tan\left(\dfrac{x}{2}\right) \] Since \(\tan(0) = 0\) and tangent is continuous at \(0\), we have: \[ \lim_{x \to 0} \tan\left(\dfrac{x}{2}\right) = \tan(0) = 0 \] **Answer:** 0 |
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| GPT-4o | To find the limit of \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we need to express the cosecant and cotangent functions in terms of sine and cosine. Recall the definitions: \[ \csc(x) = \frac{1}{\sin(x)} \] \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] Thus, we can rewrite the expression \(\csc(x) - \cot(x)\) as: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} \] Combining these into a single fraction gives: \[ \csc(x) - \cot(x) = \frac{1 - \cos(x)}{\sin(x)} \] Now, we need to find the limit as \(x\) approaches 0: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] To evaluate this limit, we can use L'Hôpital's Rule because both the numerator and the denominator approach 0 as \(x\) approaches 0. L'Hôpital's Rule states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] Here, let \(f(x) = 1 - \cos(x)\) and \(g(x) = \sin(x)\). The derivatives are: \[ f'(x) = \sin(x) \] \[ g'(x) = \cos(x) \] Applying L'Hôpital's Rule, we get: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} \] As \(x \to 0\), \(\sin(x) \to 0\) and \(\cos(x) \to 1\), so: \[ \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \frac{0}{1} = 0 \] Therefore, the solution is: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = 0 \] |
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| llama3.1-70B-Instruct | To find the limit of (csc(x) - cot(x)) as x approaches 0, we can use the following steps: 1. Recall the definitions of csc(x) and cot(x): csc(x) = 1/sin(x) cot(x) = cos(x)/sin(x) 2. Rewrite the expression (csc(x) - cot(x)) in terms of sin(x) and cos(x): csc(x) - cot(x) = 1/sin(x) - cos(x)/sin(x) 3. Simplify the expression: 1/sin(x) - cos(x)/sin(x) = (1 - cos(x))/sin(x) 4. Use the fact that sin(x) approaches 0 as x approaches 0, and the fact that cos(x) approaches 1 as x approaches 0: lim_{x \to 0} (1 - cos(x))/sin(x) = (1 - 1)/0 = 0/0 5. Use L'Hopital's rule to evaluate the limit: lim_{x \to 0} (1 - cos(x))/sin(x) = lim_{x \to 0} (-sin(x))/cos(x) 6. Simplify the expression: lim_{x \to 0} (-sin(x))/cos(x) = -1 Therefore, the limit of (csc(x) - cot(x)) as x approaches 0 is -1. |
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| llama3.1-405B-Instruct | To find the limit of \lim_{x \to 0} (csc(x) - cot(x)), we can use the following steps: 1. Recall the definitions of csc(x) and cot(x): csc(x) = 1/sin(x) cot(x) = cos(x)/sin(x) 2. Rewrite the expression in terms of sin(x) and cos(x): \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} (1/sin(x) - cos(x)/sin(x)) 3. Combine the fractions: \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} (1 - cos(x))/sin(x) 4. Use the fact that \lim_{x \to 0} sin(x) = 0 and \lim_{x \to 0} cos(x) = 1: \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} (1 - 1)/0 5. Simplify the expression: \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} 0/0 6. Use L'Hopital's rule to evaluate the limit: \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} (0 - (-sin(x)))/(cos(x)) 7. Simplify the expression: \lim_{x \to 0} (csc(x) - cot(x)) = \lim_{x \to 0} sin(x)/cos(x) 8. Use the fact that \lim_{x \to 0} sin(x) = 0 and \lim_{x \to 0} cos(x) = 1: \lim_{x \to 0} (csc(x) - cot(x)) = 0/1 9. Simplify the expression: \lim_{x \to 0} (csc(x) - cot(x)) = 0 Therefore, the limit of \lim_{x \to 0} (csc(x) - cot(x)) is 0. |
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| DeepSeek-Coder-V2 | To solve the problem \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we will follow these steps: 1. **Rewrite the expression using trigonometric identities:** \[ \csc(x) = \frac{1}{\sin(x)} \quad \text{and} \quad \cot(x) = \frac{\cos(x)}{\sin(x)} \] Therefore, \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} \] 2. **Combine the terms over a common denominator:** \[ \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] 3. **Analyze the limit of the new expression:** \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] 4. **Use the small-angle approximations for sine and cosine:** For \(x\) close to 0, \[ \sin(x) \approx x \quad \text{and} \quad \cos(x) \approx 1 - \frac{x^2}{2} \] 5. **Substitute these approximations into the expression:** \[ \frac{1 - \cos(x)}{\sin(x)} \approx \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x} = \frac{\frac{x^2}{2}}{x} = \frac{x}{2} \] 6. **Evaluate the limit of the simplified expression:** \[ \lim_{x \to 0} \frac{x}{2} = 0 \] Therefore, the solution to the problem is: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = 0 \] |
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| Qwen2-72B-Instruct | To solve the limit \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we can rewrite the expression using the definitions of \(\csc(x)\) and \(\cot(x)\): \[ \csc(x) = \frac{1}{\sin(x)} \] \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] So, \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = \lim_{x \to 0} \left( \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} \right) \] \[ = \lim_{x \to 0} \left( \frac{1 - \cos(x)}{\sin(x)} \right) \] To evaluate this limit, we can use L'Hôpital's rule since the limit takes the form \(\frac{0}{0}\) as \(x \to 0\). L'Hôpital's rule states that if the limit of the ratio of the derivatives of the numerator and denominator exists, then this limit is equal to the limit of the original ratio. Differentiating the numerator and the denominator with respect to \(x\), we get: Numerator: \(\frac{d}{dx} (1 - \cos(x)) = \sin(x)\) Denominator: \(\frac{d}{dx} \sin(x) = \cos(x)\) So, \[ \lim_{x \to 0} \left( \frac{1 - \cos(x)}{\sin(x)} \right) = \lim_{x \to 0} \left( \frac{\sin(x)}{\cos(x)} \right) \] \[ = \lim_{x \to 0} \tan(x) \] Since \(\tan(0) = 0\), \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = 0 \] Therefore, the solution to the given problem is \(\boxed{0}\). |
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| baichuan4 | To find the limit $\lim_{x \to 0} (\csc(x) - \cot(x))$, we can use L'Hôpital's Rule since the function approaches an indeterminate form of $\frac{0}{0}$ as $x$ approaches $0$. Let's start by setting up the limit: $$\lim_{x \to 0} (\csc(x) - \cot(x)) = \lim_{x \to 0} \left(\frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)}\right) = \lim_{x \to 0} \left(\frac{1 - \cos(x)}{\sin(x)}\right)$$ Now, applying L'Hôpital's Rule, we differentiate the numerator and the denominator separately: $$\lim_{x \to 0} \left(\frac{0 - (-\sin(x))}{\cos(x)}\right) = \lim_{x \to 0} \left(\frac{\sin(x)}{\cos(x)}\right)$$ This simplifies to: $$\lim_{x \to 0} (\tan(x))$$ Since $\tan(0) = 0$, the limit as $x$ approaches $0$ is: $$\lim_{x \to 0} (\tan(x)) = 0$$ Therefore, the original limit is: $$\lim_{x \to 0} (\csc(x) - \cot(x)) = 0$$ |
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| doubao | We first rewrite the expression: \[ \begin{align*} \csc(x)-\cot(x)&=\frac{1}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\ &=\frac{1-\cos(x)}{\sin(x)} \end{align*} \] Now multiply the numerator and denominator by \(1+\cos(x)\): \[ \begin{align*} \frac{1-\cos(x)}{\sin(x)}\times\frac{1+\cos(x)}{1+\cos(x)}&=\frac{1-\cos^2(x)}{\sin(x)(1+\cos(x))}\\ &=\frac{\sin^2(x)}{\sin(x)(1+\cos(x))}\\ &=\frac{\sin(x)}{1+\cos(x)} \end{align*} \] Taking the limit as \(x\to 0\): \[ \lim_{x\to 0}\frac{\sin(x)}{1+\cos(x)}=\frac{\sin(0)}{1+\cos(0)}=\frac{0}{1+1}=0 \] So the answer is 0. |
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| minimax | To find the limit \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we can use the trigonometric identities to simplify the expression. Recall that \(\csc(x) = \frac{1}{\sin(x)}\) and \(\cot(x) = \frac{\cos(x)}{\sin(x)}\). So we have: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = \lim_{x \to 0} \left(\frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)}\right) \] Now, let's combine the terms over a common denominator: \[ \lim_{x \to 0} \left(\frac{1 - \cos(x)}{\sin(x)}\right) \] We can now apply the limit directly to the numerator, since the denominator is nonzero for \(x\) approaching \(0\): \[ \lim_{x \to 0} \left(\frac{1 - \cos(x)}{\sin(x)}\right) = \frac{1 - \cos(0)}{\sin(0)} \] Since \(\cos(0) = 1\) and \(\sin(0) = 0\), we encounter a \(0/0\) indeterminate form. To resolve this, we can use L'Hôpital's Rule, which states that if we have an indeterminate form of type \(0/0\) or \(\infty/\infty\), then the limit of the functions is the same as the limit of their derivatives: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} = \lim_{x \to 0} \frac{\frac{d}{dx}(1 - \cos(x))}{\frac{d}{dx}(\sin(x))} \] Taking the derivatives, we get: \[ \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} \] Now, we can substitute \(x = 0\) directly into the simplified expression: \[ \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0 \] Therefore, the limit is: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = 0 \] |
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| kimi | To solve the limit \(\lim_{x \to 0} (\csc(x) - \cot(x))\), we can use the definitions of cosecant and cotangent in terms of sine and cosine: \[ \csc(x) = \frac{1}{\sin(x)} \] \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] So the expression becomes: \[ \lim_{x \to 0} \left(\frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)}\right) \] Combine the terms over a common denominator: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] Now, we can use the small-angle approximations for \(\sin(x) \approx x\) and \(\cos(x) \approx 1 - \frac{x^2}{2}\) as \(x \to 0\). However, these approximations are not sufficient for this limit because they would lead to an indeterminate form \(\frac{0}{0}\). Instead, we can use the fact that \(\sin(x) \approx x - \frac{x^3}{6}\) for small \(x\), which is a higher-order approximation. Using this approximation: \[ \lim_{x \to 0} \frac{1 - (1 - \frac{x^2}{2} + o(x^3))}{x - \frac{x^3}{6} + o(x^5)} \] Simplify the numerator: \[ \lim_{x \to 0} \frac{\frac{x^2}{2} + o(x^3)}{x - \frac{x^3}{6} + o(x^5)} \] Now, divide both the numerator and the denominator by \(x\): \[ \lim_{x \to 0} \frac{\frac{x}{2} + o(x^2)}{1 - \frac{x^2}{6} + o(x^4)} \] As \(x \to 0\), the higher-order terms \(o(x^2)\) and \(o(x^4)\) become negligible, and we are left with: \[ \lim_{x \to 0} \frac{\frac{x}{2}}{1} = \frac{0}{1} = 0 \] Therefore, the limit is: \[ \boxed{0} \] |
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| wenxinyiyan | 首先，我们将 csc(x) 和 cot(x) 转换为它们的正弦和余弦的比值形式： csc(x) = 1/sin(x) cot(x) = cos(x)/sin(x) 然后，我们将这些代入原式中： lim_{x \to 0} (csc(x) - cot(x)) = lim_{x \to 0} (1/sin(x) - cos(x)/sin(x)) 接下来，我们将两个分数合并为一个分数： = lim_{x \to 0} (1 - cos(x))/sin(x) 现在，我们遇到了一个 0/0 的不确定形式，这意味着我们可以使用洛必达法则来求解这个极限。洛必达法则允许我们对分子和分母分别求导，然后再次求极限： = lim_{x \to 0} (d/dx (1 - cos(x)))/d/dx sin(x) 对分子和分母分别求导，我们得到： = lim_{x \to 0} sin(x)/cos(x) 现在，我们可以直接求极限，因为当 x 趋近于 0 时，sin(x) 趋近于 0，cos(x) 趋近于 1： = 0/1 = 0 所以，答案是 0。 |
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| zhipu | To find the limit \(\lim_{x \to 0} (\csc(x) - \cot(x))\), let's first rewrite the expression in terms of sine and cosine, since \(\csc(x)\) and \(\cot(x)\) are reciprocal and quotient identities of \(\sin(x)\) and \(\cos(x)\): \[ \csc(x) = \frac{1}{\sin(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)} \] So, the expression \(\csc(x) - \cot(x)\) becomes: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] Now, we need to find the limit of this expression as \(x\) approaches 0: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] We can use the fact that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) and \(\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}\) (the latter comes from the Taylor series expansion of \(\cos(x)\) around \(x = 0\), where \(\cos(x) \approx 1 - \frac{x^2}{2}\)). To make use of these facts, we can manipulate the expression as follows: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} = \lim_{x \to 0} \left( \frac{1 - \cos(x)}{x^2} \cdot \frac{x^2}{\sin(x)} \right) \] We know that: \[ \lim_{x \to 0} \frac{x^2}{\sin(x)} = \lim_{x \to 0} x \cdot \frac{x}{\sin(x)} = \lim_{x \to 0} x \cdot 1 = 0 \] However, we need to be careful with the manipulation. Instead, let's use the identity \(1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)\): \[ \frac{1 - \cos(x)}{\sin(x)} = \frac{2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)} = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \tan\left(\frac{x}{2}\right) \] Now, as \(x\) approaches 0, \(\frac{x}{2}\) also approaches 0, and \(\tan\left(\frac{x}{2}\right)\) approaches \(\tan(0) = 0\). Therefore, the limit is: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) = \lim_{x \to 0} \tan\left(\frac{x}{2}\right) = 0 \] So, the final answer is: \[ \boxed{0} \] |
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| gpto1 | To find the limit: \[ \lim_{x \to 0} \left( \csc(x) - \cot(x) \right) \] First, recall the definitions of cosecant and cotangent in terms of sine and cosine: \[ \csc(x) = \frac{1}{\sin(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)} \] Subtracting the two gives: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] Now, consider the limit: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] We can simplify the numerator using the identity \(1 - \cos(x) = 2 \sin^2\left(\dfrac{x}{2}\right)\): \[ \frac{1 - \cos(x)}{\sin(x)} = \frac{2 \sin^2\left(\dfrac{x}{2}\right)}{2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right)} = \frac{\sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right)} = \tan\left(\dfrac{x}{2}\right) \] Therefore, the limit becomes: \[ \lim_{x \to 0} \tan\left(\dfrac{x}{2}\right) \] Since \(\tan(0) = 0\) and tangent is continuous at \(0\), we have: \[ \lim_{x \to 0} \tan\left(\dfrac{x}{2}\right) = \tan(0) = 0 \] **Answer:** 0 |
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| deepseekv3 | Alright, I have this limit problem to solve: \[ \lim_{x \to 0} (\csc(x) - \cot(x)) \] First, I need to recall what \(\csc(x)\) and \(\cot(x)\) are in terms of sine and cosine, since those are more familiar functions. I remember that: \[ \csc(x) = \frac{1}{\sin(x)} \] and \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] So, I can rewrite the expression inside the limit as: \[ \csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} \] Now, the limit becomes: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] I know that as \(x\) approaches 0, both the numerator \(1 - \cos(x)\) and the denominator \(\sin(x)\) approach 0. This is an indeterminate form of type \(\frac{0}{0}\), so I can try to use L'Hôpital's Rule, which says that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] provided that the limit on the right exists. Let's apply L'Hôpital's Rule here. First, find the derivatives of the numerator and the denominator. The derivative of the numerator \(1 - \cos(x)\) is: \[ \frac{d}{dx}(1 - \cos(x)) = 0 - (-\sin(x)) = \sin(x) \] The derivative of the denominator \(\sin(x)\) is: \[ \frac{d}{dx}(\sin(x)) = \cos(x) \] So, applying L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \lim_{x \to 0} \tan(x) \] Now, I know that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), and as \(x\) approaches 0, \(\sin(x)\) approaches 0 and \(\cos(x)\) approaches 1. Therefore: \[ \lim_{x \to 0} \tan(x) = \lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \frac{0}{1} = 0 \] So, the limit is 0. But wait, I recall that sometimes trigonometric limits can be evaluated using standard limits without resorting to L'Hôpital's Rule. Let's see if that's possible here. I remember that: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \] and \[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \] Perhaps I can manipulate the expression to use these known limits. Starting again with: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} \] I can write \(\sin(x)\) as \(x \cdot \frac{\sin(x)}{x}\), so: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{x \cdot \frac{\sin(x)}{x}} = \lim_{x \to 0} \frac{1 - \cos(x)}{x} \cdot \frac{1}{\frac{\sin(x)}{x}} \] Now, as \(x \to 0\), \(\frac{\sin(x)}{x} \to 1\), so: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{x} \cdot \frac{1}{1} = \lim_{x \to 0} \frac{1 - \cos(x)}{x} \] But I know that: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \] so: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{x} = \lim_{x \to 0} \left( \frac{1 - \cos(x)}{x^2} \cdot x \right) = \frac{1}{2} \cdot 0 = 0 \] Therefore: \[ \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)} = 0 \] This matches the result I got using L'Hôpital's Rule. So, both methods lead to the same conclusion. Finally, the limit is: \[ \boxed{0} \] |
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