| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the limit of \((\sin(2x) / x)^{1+x}\) as \(x\) approaches 0, we can use L'Hôpital's rule and properties of limits. However, L'Hôpital's rule directly applies to indeterminate forms like \(0/0\) or \(\infty/\infty\). The given function is not in a form that L'Hôpital's rule can be applied directly. Instead, we can take the natural logarithm of the function and then apply L'Hôpital's rule to the transformed expression. Let \(y = (\sin(2x) / x)^{1+x}\), then we have: \[ \ln(y) = (1+x)\ln\left(\frac{\sin(2x)}{x}\right) \] As \(x\) approaches 0, \(\sin(2x) / x\) approaches 2 (by the well-known limit \(\lim_{x\to 0}\frac{\sin(x)}{x} = 1\)). However, we need to find the limit of the entire expression. Let's focus on the limit of \(\ln(y)\): \[ \lim_{x\to 0} \ln(y) = \lim_{x\to 0} (1+x)\ln\left(\frac{\sin(2x)}{x}\right) \] This simplifies to: \[ \lim_{x\to 0} (1+x)\ln\left(\frac{\sin(2x)}{x}\right) = \lim_{x\to 0} \ln\left(\left(\frac{\sin(2x)}{x}\right)^{1+x}\right) \] Now, observe that as \(x\) approaches 0, \(\frac{\sin(2x)}{x}\) approaches 2, so we are looking at: \[ \lim_{x\to 0} \ln\left(2^{1+x}\right) \] Since \(\ln(a^b) = b\ln(a)\), we can rewrite this as: \[ \lim_{x\to 0} (1+x)\ln(2) \] This limit is straightforward, as it is simply: \[ \ln(2) \lim_{x\to 0} (1+x) = \ln(2) \cdot 1 = \ln(2) \] Therefore, the original limit of \(y\) as \(x\) approaches 0 is: \[ \lim_{x\to 0} y = e^{\ln(2)} = 2 \] So, \((\sin(2x) / x)^{1+x}\) approaches 2 as \(x\) approaches 0. |
2.0 |
| deepseekr1 | To find the limit of \(\left( \frac{\sin(2x)}{x} \right)^{1+x}\) as \(x\) approaches 0, we can use the following steps: 1. **Recognize the base and exponent behavior**: - The base \(\frac{\sin(2x)}{x}\) approaches 2 as \(x\) approaches 0. This is because \(\sin(2x)\) can be approximated by its Taylor series expansion around 0: \(\sin(2x) = 2x - \frac{(2x)^3}{6} + \cdots\), so \(\frac{\sin(2x)}{x} \approx 2 - \frac{4x^2}{3} + \cdots\). - The exponent \(1 + x\) approaches 1 as \(x\) approaches 0. 2. **Consider the logarithm of the expression**: - Let \(L\) be the limit we want to find: \(L = \lim_{x \to 0} \left( \frac{\sin(2x)}{x} \right)^{1+x}\). - Taking the natural logarithm: \(\ln L = \lim_{x \to 0} (1 + x) \ln \left( \frac{\sin(2x)}{x} \right)\). 3. **Expand the logarithm term**: - Using the Taylor series expansion for \(\ln(1 + y)\) where \(y\) is small, we get: \(\ln \left( \frac{\sin(2x)}{x} \right) \approx \ln(2) - \frac{2x^2}{3}\). 4. **Evaluate the limit of the logarithm expression**: - Substitute the expansion into the limit: \(\ln L = \lim_{x \to 0} (1 + x) \left( \ln(2) - \frac{2x^2}{3} \right)\). - Expanding this product and taking the limit as \(x \to 0\), the higher-order terms vanish, leaving \(\ln L = \ln(2)\). 5. **Exponentiate the result**: - Therefore, \(L = e^{\ln(2)} = 2\). 6. **Numerical verification**: - For very small \(x\), the expression \(\left( \frac{\sin(2x)}{x} \right)^{1+x}\) is approximated and found to be close to 2. Thus, the limit is \(\boxed{2}\). |
2.0 |
| gpto1 | To evaluate the limit \[ L = \lim_{x \to 0} \left( \frac{\sin(2x)}{x} \right)^{1 + x}, \] we start by taking the natural logarithm of \(L\) to simplify the expression: \[ \ln L = \lim_{x \to 0} (1 + x) \ln\left( \frac{\sin(2x)}{x} \right). \] Our next step is to expand \(\ln\left( \dfrac{\sin(2x)}{x} \right)\) using series expansions around \(x = 0\): 1. **Expand \(\sin(2x)\):** Using the Taylor series expansion of \(\sin(2x)\) around \(x = 0\): \[ \sin(2x) = 2x - \frac{(2x)^3}{6} + \cdots = 2x - \frac{8x^3}{6} + \cdots \] 2. **Compute \(\ln(\sin(2x))\):** We approximate \(\sin(2x)\) for small \(x\): \[ \sin(2x) \approx 2x \left(1 - \frac{4x^2}{6}\right) = 2x \left(1 - \frac{2x^2}{3}\right). \] So, \[ \ln(\sin(2x)) \approx \ln\left(2x \left(1 - \frac{2x^2}{3}\right)\right) = \ln(2x) + \ln\left(1 - \frac{2x^2}{3}\right). \] Using the approximation \(\ln(1 + y) \approx y\) for small \(y\): \[ \ln\left(1 - \frac{2x^2}{3}\right) \approx -\frac{2x^2}{3}. \] Therefore, \[ \ln(\sin(2x)) \approx \ln(2x) - \frac{2x^2}{3}. \] 3. **Compute \(\ln\left( \dfrac{\sin(2x)}{x} \right)\):** \[ \ln\left( \frac{\sin(2x)}{x} \right) = \ln(\sin(2x)) - \ln(x) \approx \ln(2x) - \frac{2x^2}{3} - \ln(x) = \ln(2) - \frac{2x^2}{3}. \] Here, we use the fact that \(\ln(2x) - \ln(x) = \ln(2)\). 4. **Compute \(\ln L\):** Now, multiply by \(1 + x\): \[ \ln L \approx (1 + x)\left[\ln(2) - \frac{2x^2}{3}\right] = \ln(2)(1 + x) - \frac{2x^2}{3}(1 + x). \] Expanding: \[ \ln L \approx \ln(2) + \ln(2) x - \frac{2x^2}{3} - \frac{2x^3}{3}. \] As \(x \to 0\), the terms involving \(x\), \(x^2\), and \(x^3\) will vanish: \[ \ln L \to \ln(2). \] 5. **Compute the limit \(L\):** Since \(\ln L \to \ln(2)\), it follows that: \[ L \to e^{\ln(2)} = 2. \] **Answer: 2** |
2.0 |
| GPT-4o | To find the limit of \(\left( \frac{\sin(2x)}{x} \right)^{1+x}\) as \(x\) approaches 0, we can break down the problem into more manageable parts. First, consider the expression inside the parentheses: \[ \frac{\sin(2x)}{x} \] As \(x\) approaches 0, we know from the standard limit that: \[ \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \] Multiplying both the numerator and the denominator by 2 gives: \[ \frac{\sin(2x)}{x} = 2 \cdot \frac{\sin(2x)}{2x} \] So: \[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} = 2 \cdot 1 = 2 \] Thus, we have: \[ \frac{\sin(2x)}{x} \to 2 \text{ as } x \to 0 \] Next, we need to consider the whole expression: \[ \left( \frac{\sin(2x)}{x} \right)^{1+x} \] We can rewrite this as: \[ \left( \frac{\sin(2x)}{x} \right)^{1+x} = 2^{1+x} \] Now we need to find the limit of \(2^{1+x}\) as \(x\) approaches 0. This is straightforward because the function \(2^{1+x}\) is continuous: \[ \lim_{x \to 0} 2^{1+x} = 2^{1+0} = 2^1 = 2 \] Therefore, the solution is: \[ \lim_{x \to 0} \left( \frac{\sin(2x)}{x} \right)^{1+x} = 2 \] |
2.0 |
| llama3.1-70B-Instruct | A classic limit problem! We can solve this using L'Hopital's rule and some trigonometric identities. First, let's rewrite the expression as: $$\left(\frac{\sin(2x)}{x}\right)^{1+x} = \exp\left((1+x)\log\left(\frac{\sin(2x)}{x}\right)\right)$$ Now, we can focus on the logarithmic part: $$\log\left(\frac{\sin(2x)}{x}\right) = \log(\sin(2x)) - \log(x)$$ As $x$ approaches 0, we have: $$\lim_{x\to 0} \log(\sin(2x)) = \log(0) = -\infty$$ and $$\lim_{x\to 0} \log(x) = -\infty$$ So, we have an indeterminate form of type $\infty - \infty$. We can apply L'Hopital's rule: $$\lim_{x\to 0} \frac{\log(\sin(2x)) - \log(x)}{1} = \lim_{x\to 0} \frac{\frac{2\cos(2x)}{\sin(2x)} - \frac{1}{x}}{1}$$ Now, we can simplify: $$\lim_{x\to 0} \frac{2\cos(2x)}{\sin(2x)} - \frac{1}{x} = \lim_{x\to 0} \frac{2\cos(2x)x - \sin(2x)}{x\sin(2x)}$$ Using the fact that $\sin(2x) \sim 2x$ and $\cos(2x) \sim 1$ as $x$ approaches 0, we get: $$\lim_{x\to 0} \frac{2x - 2x}{2x^2} = 0$$ So, we have: $$\lim_{x\to 0} \log\left(\frac{\sin(2x)}{x}\right) = 0$$ Finally, we can plug this back into the original expression: $$\lim_{x\to 0} \left(\frac{\sin(2x)}{x}\right)^{1+x} = \exp(0) = \boxed{1}$$ Therefore, the limit is 1. |
2.0 |
| llama3.1-405B-Instruct | To find the limit of (sin(2x) / x)^(1+x) as x approaches 0, we can use the following steps: 1. Use the fact that sin(2x) / x = 2 sin(x) / x as x approaches 0. 2. Use the fact that sin(x) / x approaches 1 as x approaches 0. 3. Use the fact that (1+x) approaches 1 as x approaches 0. Therefore, we have: lim (x→0) (sin(2x) / x)^(1+x) = lim (x→0) (2 sin(x) / x)^(1+x) = lim (x→0) (2 sin(x) / x)^1 = lim (x→0) 2 sin(x) / x = 2 So, the limit of (sin(2x) / x)^(1+x) as x approaches 0 is 2. |
2.0 |
| DeepSeek-Coder-V2 | To solve the problem of finding the limit of \(( \sin(2x) / x )^{1+x}\) as \(x\) approaches 0, we will follow a step-by-step approach using L'Hopital's rule and some algebraic manipulations. 1. **Rewrite the expression:** \[ \left( \frac{\sin(2x)}{x} \right)^{1+x} \] 2. **Take the natural logarithm of the expression:** \[ \ln \left( \left( \frac{\sin(2x)}{x} \right)^{1+x} \right) = (1+x) \ln \left( \frac{\sin(2x)}{x} \right) \] 3. **Simplify the logarithm:** \[ (1+x) \ln \left( \frac{\sin(2x)}{x} \right) = (1+x) \left( \ln(\sin(2x)) - \ln(x) \right) \] 4. **Evaluate the limit of the logarithm as \(x\) approaches 0:** \[ \lim_{x \to 0} (1+x) \left( \ln(\sin(2x)) - \ln(x) \right) \] 5. **Separate the terms:** \[ \lim_{x \to 0} (1+x) \ln(\sin(2x)) - \lim_{x \to 0} (1+x) \ln(x) \] 6. **Evaluate each limit separately:** - For \(\lim_{x \to 0} (1+x) \ln(\sin(2x))\): \[ \lim_{x \to 0} \ln(\sin(2x)) = \ln(0^+) \to -\infty \] Since \(1+x \to 1\) as \(x \to 0\), this term goes to \(-\infty\). - For \(\lim_{x \to 0} (1+x) \ln(x)\): \[ \lim_{x \to 0} \ln(x) = -\infty \] Since \(1+x \to 1\) as \(x \to 0\), this term also goes to \(-\infty\). 7. **Combine the results:** \[ \lim_{x \to 0} (1+x) \left( \ln(\sin(2x)) - \ln(x) \right) = -\infty - (-\infty) = 0 \] 8. **Exponentiate back to the original expression:** \[ \exp\left( \lim_{x \to 0} (1+x) \left( \ln(\sin(2x)) - \ln(x) \right) \right) = \exp(0) = 1 \] Therefore, the limit of \(( \sin(2x) / x )^{1+x}\) as \(x\) approaches 0 is: \[ \boxed{1} \] |
2.0 |
| Qwen2-72B-Instruct | To find the limit of \((\sin(2x) / x)^{1+x}\) as \(x\) approaches 0, we can use L'Hôpital's rule and properties of limits. However, L'Hôpital's rule directly applies to indeterminate forms like \(0/0\) or \(\infty/\infty\). The given function is not in a form that L'Hôpital's rule can be applied directly. Instead, we can take the natural logarithm of the function and then apply L'Hôpital's rule to the transformed expression. Let \(y = (\sin(2x) / x)^{1+x}\), then we have: \[ \ln(y) = (1+x)\ln\left(\frac{\sin(2x)}{x}\right) \] As \(x\) approaches 0, \(\sin(2x) / x\) approaches 2 (by the well-known limit \(\lim_{x\to 0}\frac{\sin(x)}{x} = 1\)). However, we need to find the limit of the entire expression. Let's focus on the limit of \(\ln(y)\): \[ \lim_{x\to 0} \ln(y) = \lim_{x\to 0} (1+x)\ln\left(\frac{\sin(2x)}{x}\right) \] This simplifies to: \[ \lim_{x\to 0} (1+x)\ln\left(\frac{\sin(2x)}{x}\right) = \lim_{x\to 0} \ln\left(\left(\frac{\sin(2x)}{x}\right)^{1+x}\right) \] Now, observe that as \(x\) approaches 0, \(\frac{\sin(2x)}{x}\) approaches 2, so we are looking at: \[ \lim_{x\to 0} \ln\left(2^{1+x}\right) \] Since \(\ln(a^b) = b\ln(a)\), we can rewrite this as: \[ \lim_{x\to 0} (1+x)\ln(2) \] This limit is straightforward, as it is simply: \[ \ln(2) \lim_{x\to 0} (1+x) = \ln(2) \cdot 1 = \ln(2) \] Therefore, the original limit of \(y\) as \(x\) approaches 0 is: \[ \lim_{x\to 0} y = e^{\ln(2)} = 2 \] So, \((\sin(2x) / x)^{1+x}\) approaches 2 as \(x\) approaches 0. |
2.0 |
| baichuan4 | To find the limit of $(\frac{\sin(2x)}{x})^{1+x}$ as $x$ approaches 0, we can use L'Hôpital's Rule or recognize that this is in indeterminate form when $x=0$. However, since this is a problem that involves trigonometric functions and exponents, we can also use Taylor series expansion to solve it. First, let's expand $\sin(2x)$ in its Taylor series around $x=0$: $$\sin(2x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots$$ Now, let's divide this series by $x$: $$\frac{\sin(2x)}{x} = 2 - \frac{(2x)^2}{3!} + \frac{(2x)^4}{5!} - \cdots$$ Next, we need to evaluate $(1+x)$th power of this series. For this, we can use the binomial theorem: $$(2 - \frac{(2x)^2}{3!} + \frac{(2x)^4}{5!} - \cdots)^{1+x} = 2^{1+x}(1 - \frac{(2x)^2}{3!} + \frac{(2x)^4}{5!} - \cdots)^{1+x}$$ Now, we can expand the right-hand side using the binomial theorem again: $$2^{1+x}(1 - \frac{(2x)^2}{3!} + \frac{(2x)^4}{5!} - \cdots)^{1+x} = 2^{1+x}(1 + (1+x)(-\frac{(2x)^2}{3!}) + \frac{(1+x)(1+x-1)}{2}(-\frac{(2x)^2}{3!})^2 + \cdots)$$ As $x$ approaches 0, the terms with $x$ in the denominator will vanish, and we will be left with: $$2^{1+0}(1 + (1+0)(-\frac{(2*0)^2}{3!}) + \frac{(1+0)(1+0-1)}{2}(-\frac{(2*0)^2}{3!})^2 + \cdots) = 2(1 + 0 + 0 + \cdots) = 2$$ Therefore, the limit of $(\frac{\sin(2x)}{x})^{1+x}$ as $x$ approaches 0 is 2. |
2.0 |
| doubao | As x approaches 0, sin(2x) is approximately 2x. So we can rewrite the expression as (2x / x)^(1+x) = 2^(1+x). As x approaches 0, 1+x approaches 1, so the limit is 2^1 = 2. | 2.0 |
| minimax | To solve the limit of \((\frac{\sin(2x)}{x})^{(1+x)}\) as \(x\) approaches 0, we can use the fact that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\). However, we need to be careful with the exponent \(1+x\) as \(x\) approaches 0. Let's break down the problem into two parts: the base and the exponent. First, let's find the limit of the base \(\frac{\sin(2x)}{x}\) as \(x\) approaches 0: \[ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{2\sin(2x)}{2x} = 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} \] Now, let \(u = 2x\), so when \(x \to 0\), \(u \to 0\). Then we have: \[ 2 \cdot \lim_{u \to 0} \frac{\sin(u)}{u} = 2 \cdot 1 = 2 \] So the base \(\frac{\sin(2x)}{x}\) approaches 2 as \(x\) approaches 0. Now let's look at the exponent \(1+x\). As \(x\) approaches 0, \(1+x\) approaches 1. Therefore, we have: \[ \lim_{x \to 0} \left(\frac{\sin(2x)}{x}\right)^{(1+x)} = \lim_{x \to 0} 2^{(1+x)} \] Since the exponent \(1+x\) approaches 1 as \(x\) approaches 0, we have: \[ \lim_{x \to 0} 2^{(1+x)} = 2^1 = 2 \] So the limit of \((\frac{\sin(2x)}{x})^{(1+x)}\) as \(x\) approaches 0 is 2. |
2.0 |
| kimi | To find the limit of the expression \(\left(\frac{\sin(2x)}{x}\right)^{1+x}\) as \(x\) approaches 0, we can follow these steps: 1. **Simplify the Inner Expression**: First, we need to simplify the inner expression \(\frac{\sin(2x)}{x}\). As \(x \to 0\), we know from the small angle approximation that \(\sin(2x) \approx 2x\). Therefore, \(\frac{\sin(2x)}{x} \approx \frac{2x}{x} = 2\). However, this approximation is not exact, so we need to be careful. 2. **Use L'Hôpital's Rule**: Since we have an indeterminate form \(\frac{0}{0}\) as \(x \to 0\), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator with respect to \(x\): \[ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{\cos(2x) \cdot 2}{1} = 2 \cos(0) = 2 \] This tells us that \(\frac{\sin(2x)}{x} \to 2\) as \(x \to 0\). 3. **Evaluate the Outer Expression**: Now, we substitute this limit into the outer expression: \[ \left(\frac{\sin(2x)}{x}\right)^{1+x} \to 2^{1+x} \text{ as } x \to 0 \] 4. **Simplify the Exponent**: As \(x \to 0\), the exponent \(1+x\) also approaches 1. Therefore, we have: \[ 2^{1+x} \to 2^1 = 2 \text{ as } x \to 0 \] So, the limit of \(\left(\frac{\sin(2x)}{x}\right)^{1+x}\) as \(x\) approaches 0 is \(\boxed{2}\). |
2.0 |
| wenxinyiyan | 为了解决这个问题，我们需要使用到一些数学上的概念，包括极限、等价无穷小、泰勒展开等。 首先，我们知道当x趋近于0时，sin(x)与x是等价无穷小，即： sin(x) ~ x 那么，sin(2x)就可以表示为2x，所以原式可以写为： (sin(2x) / x)^(1+x) = (2x / x)^(1+x) = 2^(1+x) 当x趋近于0时，2^(1+x)的极限就是2^1 = 2。 所以，当x趋近于0时，(sin(2x) / x)^(1+x)的极限是2。 |
2.0 |
| zhipu | To solve the limit of the function \((\frac{\sin(2x)}{x})^{1+x}\) as \(x\) approaches 0, we need to carefully handle the indeterminate forms that arise. The main challenge here is the form \(\frac{0}{0}\) inside the parentheses as \(x\) approaches 0, and the exponent \(1+x\) also approaching 1. Let's break down the solution step by step: ### Step 1: Simplify the Expression Inside the Parentheses First, consider the expression inside the parentheses: \[ \frac{\sin(2x)}{x} \] We know that \(\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1\) because \(\sin(2x) \approx 2x\) for small \(x\). Therefore, we can rewrite the expression as: \[ \frac{\sin(2x)}{x} = \frac{\sin(2x)}{2x} \cdot 2 \] As \(x \to 0\): \[ \frac{\sin(2x)}{2x} \to 1 \implies \frac{\sin(2x)}{x} \to 2 \] ### Step 2: Consider the Exponent Now, consider the exponent \(1 + x\). As \(x \to 0\), \(1 + x \to 1\). ### Step 3: Combine the Results We now have: \[ \left(\frac{\sin(2x)}{x}\right)^{1+x} \approx 2^{1+x} \] As \(x \to 0\), \(1 + x \to 1\), so: \[ 2^{1+x} \to 2^1 = 2 \] ### Step 4: Confirm with Natural Logarithms (Optional) For a more rigorous approach, we can use natural logarithms to handle the limit. Let \(L\) be the limit we want to find: \[ L = \lim_{x \to 0} \left(\frac{\sin(2x)}{x}\right)^{1+x} \] Take the natural logarithm of both sides: \[ \ln L = \lim_{x \to 0} (1+x) \ln\left(\frac{\sin(2x)}{x}\right) \] We already know that \(\frac{\sin(2x)}{x} \to 2\) as \(x \to 0\), so: \[ \ln\left(\frac{\sin(2x)}{x}\right) \to \ln(2) \] Thus: \[ \ln L = \lim_{x \to 0} (1+x) \ln(2) = \ln(2) \lim_{x \to 0} (1+x) = \ln(2) \cdot 1 = \ln(2) \] Exponentiating both sides to solve for \(L\): \[ L = e^{\ln(2)} = 2 \] ### Final Answer Therefore, the limit of \((\frac{\sin(2x)}{x})^{1+x}\) as \(x\) approaches 0 is: \[ \boxed{2} \] |
2.0 |
| gpto1 | To evaluate the limit \[ L = \lim_{x \to 0} \left( \frac{\sin(2x)}{x} \right)^{1 + x}, \] we start by taking the natural logarithm of \(L\) to simplify the expression: \[ \ln L = \lim_{x \to 0} (1 + x) \ln\left( \frac{\sin(2x)}{x} \right). \] Our next step is to expand \(\ln\left( \dfrac{\sin(2x)}{x} \right)\) using series expansions around \(x = 0\): 1. **Expand \(\sin(2x)\):** Using the Taylor series expansion of \(\sin(2x)\) around \(x = 0\): \[ \sin(2x) = 2x - \frac{(2x)^3}{6} + \cdots = 2x - \frac{8x^3}{6} + \cdots \] 2. **Compute \(\ln(\sin(2x))\):** We approximate \(\sin(2x)\) for small \(x\): \[ \sin(2x) \approx 2x \left(1 - \frac{4x^2}{6}\right) = 2x \left(1 - \frac{2x^2}{3}\right). \] So, \[ \ln(\sin(2x)) \approx \ln\left(2x \left(1 - \frac{2x^2}{3}\right)\right) = \ln(2x) + \ln\left(1 - \frac{2x^2}{3}\right). \] Using the approximation \(\ln(1 + y) \approx y\) for small \(y\): \[ \ln\left(1 - \frac{2x^2}{3}\right) \approx -\frac{2x^2}{3}. \] Therefore, \[ \ln(\sin(2x)) \approx \ln(2x) - \frac{2x^2}{3}. \] 3. **Compute \(\ln\left( \dfrac{\sin(2x)}{x} \right)\):** \[ \ln\left( \frac{\sin(2x)}{x} \right) = \ln(\sin(2x)) - \ln(x) \approx \ln(2x) - \frac{2x^2}{3} - \ln(x) = \ln(2) - \frac{2x^2}{3}. \] Here, we use the fact that \(\ln(2x) - \ln(x) = \ln(2)\). 4. **Compute \(\ln L\):** Now, multiply by \(1 + x\): \[ \ln L \approx (1 + x)\left[\ln(2) - \frac{2x^2}{3}\right] = \ln(2)(1 + x) - \frac{2x^2}{3}(1 + x). \] Expanding: \[ \ln L \approx \ln(2) + \ln(2) x - \frac{2x^2}{3} - \frac{2x^3}{3}. \] As \(x \to 0\), the terms involving \(x\), \(x^2\), and \(x^3\) will vanish: \[ \ln L \to \ln(2). \] 5. **Compute the limit \(L\):** Since \(\ln L \to \ln(2)\), it follows that: \[ L \to e^{\ln(2)} = 2. \] **Answer: 2** |
2.0 |
| deepseekv3 | To evaluate the limit of \(\left(\frac{\sin(2x)}{x}\right)^{1+x}\) as \(x\) approaches 0, we can proceed with the following steps: ### Step 1: Simplify the Expression First, consider the expression inside the parentheses: \[ \frac{\sin(2x)}{x} \] We know the standard limit: \[ \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \] Multiplying both sides by 2, we get: \[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \] ### Step 2: Analyze the Exponent The exponent in the original expression is \(1 + x\). As \(x\) approaches 0, \(1 + x\) approaches 1. ### Step 3: Combine the Results Now, consider the entire expression: \[ \left(\frac{\sin(2x)}{x}\right)^{1+x} \] As \(x\) approaches 0, \(\frac{\sin(2x)}{x}\) approaches 2, and the exponent \(1 + x\) approaches 1. Therefore: \[ \lim_{x \to 0} \left(\frac{\sin(2x)}{x}\right)^{1+x} = 2^1 = 2 \] ### Final Answer \[ \boxed{2} \] |
2.0 |