| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to evaluate the limit of the given expression as \( r \) approaches infinity. The expression involves integrals of the form \( \int_0^{\pi/2} x^r \sin(x) dx \) and \( \int_0^{\pi/2} x^r \cos(x) dx \). Let's denote \( I_1(r) = \int_0^{\pi/2} x^r \sin(x) dx \) and \( I_2(r) = \int_0^{\pi/2} x^r \cos(x) dx \). The given expression can be rewritten as \( \lim_{r \to \infty} \frac{I_1(r)}{rI_2(r)} \). To evaluate the limit, we can use L'Hôpital's rule, which states that if the limit of the ratio of the derivatives of the numerator and denominator exists, then it is equal to the limit of the original ratio. In this case, we need to find the limit of the ratio of the derivatives of \( I_1(r) \) and \( rI_2(r) \) with respect to \( r \). Differentiating \( I_1(r) \) with respect to \( r \), we get \( I_1'(r) = \int_0^{\pi/2} x^r \ln(x) \sin(x) dx \). Similarly, differentiating \( rI_2(r) \) with respect to \( r \), we get \( (rI_2(r))' = I_2(r) + rI_2'(r) = I_2(r) + r\int_0^{\pi/2} x^r \ln(x) \cos(x) dx \). Now, we can apply L'Hôpital's rule and evaluate the limit of the ratio of the derivatives: \[ \lim_{r \to \infty} \frac{I_1'(r)}{(rI_2(r))'} = \lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) dx}{I_2(r) + r\int_0^{\pi/2} x^r \ln(x) \cos(x) dx}. \] To evaluate this limit, we need to consider the behavior of the integrals as \( r \) approaches infinity. Notice that as \( r \) increases, the integrand \( x^r \ln(x) \sin(x) \) decays rapidly to zero for \( x \in (0, \pi/2) \), while \( x^r \ln(x) \cos(x) \) also decays rapidly to zero. Therefore, both the numerator and the denominator of the limit will approach zero as \( r \) approaches infinity. In this case, we can apply L'Hôpital's rule again to evaluate the limit of the ratio of the second derivatives of \( I_1(r) \) and \( rI_2(r) \) with respect to \( r \). However, this process will not lead to a simpler expression. Instead, we can analyze the behavior of the integrals as \( r \) approaches infinity to determine the limit. As \( r \) approaches infinity, the integrals \( \int_0^{\pi/2} x^r \ln(x) \sin(x) dx \) and \( \int_0^{\pi/2} x^r \ln(x) \cos(x) dx \) will be dominated by the behavior of the integrands near \( x = 0 \). Near \( x = 0 \), we have \( \sin(x) \approx x \) and \( \cos(x) \approx 1 \), so the integrands become \( x^r \ln(x) x \) and \( x^r \ln(x) \), respectively. The integral \( \int_0^{\pi/2} x^r \ln(x) x dx \) will be dominated by the term \( x^r \ln(x) \) as \( r \) approaches infinity, while the integral \( \int_0^{\pi/2} x^r \ln(x) dx \) will also be dominated by the term \( x^r \ln(x) \). Therefore, the limit of the ratio of the derivatives of \( I_1(r) \) and \( rI_2(r) \) with respect to \( r \) will be equal to the limit of the ratio of the integrands \( x^r \ln(x) x \) and \( x^r \ln(x) \) as \( r \) approaches infinity. Since \( x^r \ln(x) x \) and \( x^r \ln(x) \) both decay to zero as \( r \) approaches infinity, we can evaluate the limit of their ratio as \( r \) approaches infinity by considering the limit of their ratio as \( x \) approaches zero. As \( x \) approaches zero, \( \ln(x) x \) approaches zero, so the limit of the ratio of the integrands will be equal to the limit of the ratio of \( x^r \) and \( x^r \), which is equal to \( 1 \). Therefore, the limit of the given expression as \( r \) approaches infinity is \( \boxed{1} \). |
0.63662 |
| deepseekr1 | To find the limit as \( r \) approaches infinity of the ratio \[ \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx}, \] we analyze the behavior of the integrals near the upper limit \( x = \pi/2 \). For large \( r \), the integrands \( x^r \sin(x) \) and \( x^r \cos(x) \) are sharply peaked near \( x = \pi/2 \). We use the substitution \( t = \pi/2 - x \) to approximate the integrals around \( t = 0 \). For the numerator integral: \[ \int_0^{\pi/2} x^r \sin(x) \, dx \approx \int_0^\infty (\pi/2 - t)^r \cos(t) \, dt \approx (\pi/2)^r \int_0^\infty e^{-rt/(\pi/2)} \cos(t) \, dt. \] For the denominator integral: \[ \int_0^{\pi/2} x^r \cos(x) \, dx \approx \int_0^\infty (\pi/2 - t)^r \sin(t) \, dt \approx (\pi/2)^r \int_0^\infty e^{-rt/(\pi/2)} \sin(t) \, dt. \] Using the substitution \( s = rt/(\pi/2) \), we convert these integrals into forms that can be evaluated for large \( r \). After substitution and simplifying, both integrals are approximated by their leading terms. The key steps involve expanding the trigonometric functions around \( t = 0 \) and using the exponential decay of the integrand. The leading terms for the integrals are found to be: \[ \int_0^{\pi/2} x^r \sin(x) \, dx \approx \frac{(\pi/2)^{r+1}}{r} \quad \text{and} \quad \int_0^{\pi/2} x^r \cos(x) \, dx \approx \frac{(\pi/2)^{r+2}}{r^2}. \] Forming the ratio and simplifying, we find: \[ \frac{\frac{(\pi/2)^{r+1}}{r}}{r \cdot \frac{(\pi/2)^{r+2}}{r^2}} = \frac{2}{\pi}. \] Thus, the limit is: \[ \boxed{\dfrac{2}{\pi}} \] |
0.63662 |
| gpto1 | To find the limit \[ L = \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin x \, dx}{r \int_0^{\pi/2} x^r \cos x \, dx}, \] we can use asymptotic analysis for large \( r \). As \( r \rightarrow \infty \), the integrals are dominated by the behavior near the upper limit \( x = \frac{\pi}{2} \) because \( x^r \) becomes significant only when \( x \) is close to \( \frac{\pi}{2} \) (since any number less than \( \frac{\pi}{2} \) raised to a large power becomes negligible). Let's make the substitution \( x = \frac{\pi}{2} - t \), with \( t \geq 0 \) small when \( x \) is near \( \frac{\pi}{2} \). Then \( dx = -dt \), and the limits of integration for \( t \) become from 0 to \( \frac{\pi}{2} \), after changing the sign due to \( dx = -dt \). We approximate the functions near \( x = \frac{\pi}{2} \): \[ x^r \approx \left( \frac{\pi}{2} - t \right)^r = \left( \frac{\pi}{2} \right)^r e^{-r \left( \frac{2t}{\pi} + \frac{2t^2}{\pi^2} \right)} \approx \left( \frac{\pi}{2} \right)^r e^{- \frac{2 r t}{\pi}}. \] \[ \sin x = \sin\left( \frac{\pi}{2} - t \right) \approx \cos t \approx 1 - \frac{t^2}{2}. \] \[ \cos x = \cos\left( \frac{\pi}{2} - t \right) \approx \sin t \approx t - \frac{t^3}{6}. \] Substituting back into the integrals: \[ \int_0^{\pi/2} x^r \sin x \, dx \approx \left( \frac{\pi}{2} \right)^r \int_0^\infty e^{- \frac{2 r t}{\pi}} \left( 1 - \frac{t^2}{2} \right) dt. \] \[ r \int_0^{\pi/2} x^r \cos x \, dx \approx r \left( \frac{\pi}{2} \right)^r \int_0^\infty e^{- \frac{2 r t}{\pi}} \left( t - \frac{t^3}{6} \right) dt. \] Evaluating the integrals using standard Laplace transforms (noting that \( e^{-a t} t^n \) integrates to \( \frac{n!}{a^{n+1}} \)): \[ \int_0^\infty e^{-a t} dt = \frac{1}{a}, \quad \int_0^\infty e^{-a t} t^n dt = \frac{n!}{a^{n+1}}. \] With \( a = \frac{2 r}{\pi} \), we find: \[ \int_0^\infty e^{- \frac{2 r t}{\pi}} dt = \frac{\pi}{2 r}, \quad \int_0^\infty e^{- \frac{2 r t}{\pi}} t^2 dt = \frac{2! \pi^3}{(2 r)^3}. \] Substituting back: \[ \int_0^{\pi/2} x^r \sin x \, dx \approx \left( \frac{\pi}{2} \right)^r \left( \frac{\pi}{2 r} - \frac{\pi^3}{(2 r)^3} \right). \] \[ r \int_0^{\pi/2} x^r \cos x \, dx \approx r \left( \frac{\pi}{2} \right)^r \left( \frac{\pi^2}{(2 r)^2} - \frac{\pi^4}{(2 r)^4} \right). \] Simplifying the ratio \( L \): \[ L = \lim_{r \rightarrow \infty} \frac{\left( \frac{\pi}{2 r} - \frac{\pi^3}{(2 r)^3} \right)}{\left( \frac{\pi^2}{(2 r)^2} - \frac{\pi^4}{(2 r)^4} \right)} = \lim_{r \rightarrow \infty} \frac{\frac{\pi}{2 r}}{\frac{\pi^2}{(2 r)^2}} = \frac{1}{\pi}, \] since the higher-order terms become negligible as \( r \) approaches infinity. **Answer:** **\(\displaystyle \frac{1}{\pi}\)** |
0.63662 |
| GPT-4o | To evaluate the limit \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx}, \] we start by analyzing the behavior of the integrals \(\int_0^{\pi/2} x^r \sin(x) \, dx\) and \(\int_0^{\pi/2} x^r \cos(x) \, dx\) for large \(r\). As \(r\) becomes very large, most of the contribution to these integrals will come from values of \(x\) close to the upper limit \(\pi/2\). To see this, we can use the method of asymptotic approximation or Laplace's method. First, consider the integral \(\int_0^{\pi/2} x^r \sin(x) \, dx\). For large \(r\), the integrand \(x^r \sin(x)\) is sharply peaked near \(x = \pi/2\). Near \(x = \pi/2\), we can approximate: \[ \sin(x) \approx \sin\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right)(x - \frac{\pi}{2}) = 1. \] Thus, for large \(r\), \[ \int_0^{\pi/2} x^r \sin(x) \, dx \approx \int_0^{\pi/2} x^r \, dx. \] Now consider the integral \(\int_0^{\pi/2} x^r \cos(x) \, dx\). Similarly, near \(x = \pi/2\), \[ \cos(x) \approx \cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)(x - \frac{\pi}{2}) = 0 - (x - \frac{\pi}{2}) = \frac{\pi}{2} - x. \] Thus, for large \(r\), \[ \int_0^{\pi/2} x^r \cos(x) \, dx \approx \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx. \] Next, we evaluate the integrals. For \( \int_0^{\pi/2} x^r \, dx \): \[ \int_0^{\pi/2} x^r \, dx = \left. \frac{x^{r+1}}{r+1} \right|_0^{\pi/2} = \frac{(\pi/2)^{r+1}}{r+1}. \] For \( \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx \): \[ \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx = \frac{\pi}{2} \int_0^{\pi/2} x^r \, dx - \int_0^{\pi/2} x^{r+1} \, dx. \] We already know \( \int_0^{\pi/2} x^r \, dx = \frac{(\pi/2)^{r+1}}{r+1} \). For \( \int_0^{\pi/2} x^{r+1} \, dx \): \[ \int_0^{\pi/2} x^{r+1} \, dx = \left. \frac{x^{r+2}}{r+2} \right|_0^{\pi/2} = \frac{(\pi/2)^{r+2}}{r+2}. \] Putting it all together: \[ \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx = \frac{\pi}{2} \cdot \frac{(\pi/2)^{r+1}}{r+1} - \frac{(\pi/2)^{r+2}}{r+2}. \] Simplifying, \[ \frac{\pi}{2} \cdot \frac{(\pi/2)^{r+1}}{r+1} - \frac{(\pi/2)^{r+2}}{r+2} = \frac{(\pi/2)^{r+2}}{r+1} - \frac{(\ |
0.63662 |
| llama3.1-70B-Instruct | ## Step 1: Recognize that the problem involves evaluating a limit that includes integrals. The problem asks for the limit as $r$ approaches infinity of a ratio involving two integrals. The first step is to understand the structure of the problem and identify the integrals involved. ## Step 2: Identify the integrals and consider possible approaches for evaluating them. The integrals are $\int_0^{\pi/2} x^r \sin(x) dx$ and $\int_0^{\pi/2} x^r \cos(x) dx$. Given the presence of $x^r$ and the trigonometric functions, direct integration may be challenging. Considering the limit as $r$ approaches infinity suggests looking for a method that simplifies the integrals or their ratio. ## Step 3: Consider using integration by parts or a similar technique to simplify the integrals. Integration by parts could be a viable method for simplifying these integrals, especially since we are dealing with a product of functions. However, given the limit and the structure of the integrals, another approach might be more suitable. ## Step 4: Recognize the potential for using the properties of limits and integrals to simplify the expression. Given the limit is on the ratio of the integrals and not the integrals themselves, we can consider the properties of limits and how they interact with integrals. Specifically, the limit of a ratio can sometimes be simplified by considering the behavior of the numerator and denominator separately. ## Step 5: Consider the behavior of the integrals as $r$ approaches infinity. As $r$ approaches infinity, the term $x^r$ in both integrals will dominate the behavior of the integrals. For $x$ in the interval $[0, \pi/2]$, $x^r$ will approach 0 for $x < 1$ and will approach infinity for $x = 1$. However, since the upper limit of integration is $\pi/2$, which is less than 1, the term $x^r$ will approach 0 for all $x$ in the interval of integration. ## Step 6: Analyze the impact of the trigonometric functions on the integrals. The presence of $\sin(x)$ and $\cos(x)$ in the integrals will not significantly affect the overall behavior of the integrals as $r$ approaches infinity, since the $x^r$ term dominates. ## Step 7: Consider the limit of the ratio of the integrals. Given that both integrals approach 0 as $r$ approaches infinity, we need to consider the limit of the ratio. This is an indeterminate form 0/0, suggesting the use of L'Hôpital's rule or a similar approach. ## Step 8: Apply L'Hôpital's rule or a similar method to evaluate the limit of the ratio. However, given the complexity of directly applying L'Hôpital's rule to the integrals, we reconsider the approach. Instead, we note that the limit can be simplified by considering the behavior of the integrands. ## Step 9: Simplify the expression by considering the behavior of the integrands as $r$ approaches infinity. As $r$ approaches infinity, the term $x^r$ dominates the behavior of the integrands. The ratio of the integrals can be simplified by considering the limit of the ratio of the integrands. ## Step 10: Evaluate the limit of the ratio of the integrands. The limit of the ratio of the integrands as $r$ approaches infinity can be evaluated by considering the behavior of $x^r \sin(x)$ and $x^r \cos(x)$. ## Step 11: Simplify the expression by canceling out the $x^r$ term. The $x^r$ term can be canceled out from the numerator and denominator, leaving the limit of the ratio of $\sin(x)$ and $\cos(x)$. ## Step 12: Evaluate the limit of the ratio of $\sin(x)$ and $\cos(x)$. The limit of the ratio of $\sin(x)$ and $\cos(x)$ as $x$ approaches $\pi/2$ is 1. ## Step 13: Consider the effect of the $r$ term in the denominator of the original expression. The original expression includes a term $r$ in the denominator. As $r$ approaches infinity, this term will dominate the behavior of the expression. ## Step 14: Evaluate the final limit. The final limit is the limit of the ratio of the integrals, which is equal to the limit of the ratio of the integrands, multiplied by the limit of $1/r$. The final answer is: $\boxed{0}$ |
0.63662 |
| llama3.1-405B-Instruct | To evaluate the limit, we can use the following steps: 1. Use integration by parts to evaluate the integrals: $$\int_0^{\pi/2} x^r \sin(x) dx = -x^r \cos(x) \Big|_0^{\pi/2} + r \int_0^{\pi/2} x^{r-1} \cos(x) dx$$ $$\int_0^{\pi/2} x^r \cos(x) dx = x^r \sin(x) \Big|_0^{\pi/2} - r \int_0^{\pi/2} x^{r-1} \sin(x) dx$$ 2. Simplify the expressions: $$\int_0^{\pi/2} x^r \sin(x) dx = -\left(\frac{\pi}{2}\right)^r + r \int_0^{\pi/2} x^{r-1} \cos(x) dx$$ $$\int_0^{\pi/2} x^r \cos(x) dx = \left(\frac{\pi}{2}\right)^r - r \int_0^{\pi/2} x^{r-1} \sin(x) dx$$ 3. Substitute the expressions into the original limit: $$\lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \sin(x) dx}{r \int_0^{\pi/2} x^r \cos(x) dx} = \lim_{r \to \infty} \frac{-\left(\frac{\pi}{2}\right)^r + r \int_0^{\pi/2} x^{r-1} \cos(x) dx}{r \left(\left(\frac{\pi}{2}\right)^r - r \int_0^{\pi/2} x^{r-1} \sin(x) dx\right)}$$ 4. Simplify the expression: $$\lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \sin(x) dx}{r \int_0^{\pi/2} x^r \cos(x) dx} = \lim_{r \to \infty} \frac{-\left(\frac{\pi}{2}\right)^r + r \int_0^{\pi/2} x^{r-1} \cos(x) dx}{r \left(\frac{\pi}{2}\right)^r - r^2 \int_0^{\pi/2} x^{r-1} \sin(x) dx}$$ 5. Evaluate the limit: $$\lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \sin(x) dx}{r \int_0^{\pi/2} x^r \cos(x) dx} = \lim_{r \to \infty} \frac{-\left(\frac{\pi}{2}\right)^r + r \int_0^{\pi/2} x^{r-1} \cos(x) dx}{r \left(\frac{\pi}{2}\right)^r - r^2 \int_0^{\pi/2} x^{r-1} \sin(x) dx} = \frac{1}{\pi}$$ Therefore, the final answer is $\boxed{\frac{1}{\pi}}$. |
0.63662 |
| DeepSeek-Coder-V2 | To solve the given problem, we need to find the limit: \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx} \] We will use L'Hopital's rule, which is applicable when we have an indeterminate form of the type \(\frac{\infty}{\infty}\). To apply L'Hopital's rule, we need to differentiate the numerator and the denominator with respect to \(r\). First, let's define the functions: \[ f(r) = \int_0^{\pi/2} x^r \sin(x) \, dx \] \[ g(r) = r \int_0^{\pi/2} x^r \cos(x) \, dx \] We need to find the derivatives \(f'(r)\) and \(g'(r)\). ### Differentiating \(f(r)\): Using Leibniz's rule for differentiating under the integral sign: \[ f'(r) = \frac{d}{dr} \left( \int_0^{\pi/2} x^r \sin(x) \, dx \right) = \int_0^{\pi/2} \frac{\partial}{\partial r} (x^r \sin(x)) \, dx \] Since \(x^r = e^{r \ln(x)}\), we have: \[ \frac{\partial}{\partial r} (x^r \sin(x)) = x^r \ln(x) \sin(x) \] Thus, \[ f'(r) = \int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx \] ### Differentiating \(g(r)\): Using the product rule and Leibniz's rule: \[ g(r) = r \int_0^{\pi/2} x^r \cos(x) \, dx \] \[ g'(r) = \frac{d}{dr} \left( r \int_0^{\pi/2} x^r \cos(x) \, dx \right) = \int_0^{\pi/2} x^r \cos(x) \, dx + r \int_0^{\pi/2} \frac{\partial}{\partial r} (x^r \cos(x)) \, dx \] Since \(x^r = e^{r \ln(x)}\), we have: \[ \frac{\partial}{\partial r} (x^r \cos(x)) = x^r \ln(x) \cos(x) \] Thus, \[ g'(r) = \int_0^{\pi/2} x^r \cos(x) \, dx + r \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx \] ### Applying L'Hopital's rule: \[ \lim_{r \rightarrow \infty} \frac{f(r)}{g(r)} = \lim_{r \rightarrow \infty} \frac{f'(r)}{g'(r)} \] Substituting the derivatives we found: \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{\int_0^{\pi/2} x^r \cos(x) \, dx + r \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx} \] As \(r \rightarrow \infty\), the dominant term in the denominator is \(r \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx\) because \(x^r \ln(x)\) grows faster than \(x^r\). Therefore, the ratio simplifies to: \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{r \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx} \] Since both the numerator and the denominator are integrals of similar forms, we can simplify further: \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{r \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx} = \lim_{r \rightarrow \infty} \frac{\sin(x)}{r \cos(x)} = 0 \] Thus, the final answer is: \[ \boxed{0} \] |
0.63662 |
| Qwen2-72B-Instruct | To solve this problem, we need to evaluate the limit of the given expression as $r$ approaches infinity. The expression involves integrals of the form $\int_0^{\pi/2} x^r \sin(x) dx$ and $\int_0^{\pi/2} x^r \cos(x) dx$. Let's denote $I_1(r) = \int_0^{\pi/2} x^r \sin(x) dx$ and $I_2(r) = \int_0^{\pi/2} x^r \cos(x) dx$. The given expression can be rewritten as $\lim_{r \to \infty} \frac{I_1(r)}{rI_2(r)}$. To evaluate the limit, we can use L'Hôpital's rule, which states that if the limit of the ratio of the derivatives of the numerator and denominator exists, then it is equal to the limit of the original ratio. In this case, we need to find the limit of the ratio of the derivatives of $I_1(r)$ and $rI_2(r)$ with respect to $r$. Differentiating $I_1(r)$ with respect to $r$, we get $I_1'(r) = \int_0^{\pi/2} x^r \ln(x) \sin(x) dx$. Similarly, differentiating $rI_2(r)$ with respect to $r$, we get $(rI_2(r))' = I_2(r) + rI_2'(r) = I_2(r) + r\int_0^{\pi/2} x^r \ln(x) \cos(x) dx$. Now, we can apply L'Hôpital's rule and evaluate the limit of the ratio of the derivatives: \[ \lim_{r \to \infty} \frac{I_1'(r)}{(rI_2(r))'} = \lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) dx}{I_2(r) + r\int_0^{\pi/2} x^r \ln(x) \cos(x) dx}. \] To evaluate this limit, we need to consider the behavior of the integrals as $r$ approaches infinity. Notice that as $r$ increases, the integrand $x^r \ln(x) \sin(x)$ decays rapidly to zero for $x \in (0, \pi/2)$, while $x^r \ln(x) \cos(x)$ also decays rapidly to zero. Therefore, both the numerator and the denominator of the limit will approach zero as $r$ approaches infinity. In this case, we can apply L'Hôpital's rule again to evaluate the limit of the ratio of the second derivatives of $I_1(r)$ and $rI_2(r)$ with respect to $r$. However, this process will not lead to a simpler expression. Instead, we can analyze the behavior of the integrals as $r$ approaches infinity to determine the limit. As $r$ approaches infinity, the integrals $\int_0^{\pi/2} x^r \ln(x) \sin(x) dx$ and $\int_0^{\pi/2} x^r \ln(x) \cos(x) dx$ will be dominated by the behavior of the integrands near $x = 0$. Near $x = 0$, we have $\sin(x) \approx x$ and $\cos(x) \approx 1$, so the integrands become $x^r \ln(x) x$ and $x^r \ln(x)$, respectively. The integral $\int_0^{\pi/2} x^r \ln(x) x dx$ will be dominated by the term $x^r \ln(x)$ as $r$ approaches infinity, while the integral $\int_0^{\pi/2} x^r \ln(x) dx$ will also be dominated by the term $x^r \ln(x)$. Therefore, the limit of the ratio of the derivatives of $I_1(r)$ and $rI_2(r)$ with respect to $r$ will be equal to the limit of the ratio of the integrands $x^r \ln(x) x$ and $x^r \ln(x)$ as $r$ approaches infinity. Since $x^r \ln(x) x$ and $x^r \ln(x)$ both decay to zero as $r$ approaches infinity, we can evaluate the limit of their ratio as $r$ approaches infinity by considering the limit of their ratio as $x$ approaches zero. As $x$ approaches zero, $\ln(x) x$ approaches zero, so the limit of the ratio of the integrands will be equal to the limit of the ratio of $x^r$ and $x^r$, which is equal to $1$. Therefore, the limit of the given expression as $r$ approaches infinity is $\boxed{1}$. |
0.63662 |
| baichuan4 | To solve the limit problem, we can analyze the integrals involved and apply L'Hôpital's rule for indeterminate forms. The given limit is of the form $\frac{0}{0}$ as $r \rightarrow \infty$, which allows us to differentiate the numerator and the denominator with respect to $r$. Let's denote the given limit as $L$: $$L = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r\int_0^{\pi/2} x^r \cos(x) \, dx}$$ We will differentiate the numerator and the denominator with respect to $r$. By the Leibniz integral rule, we can differentiate under the integral sign: $$\frac{d}{dr} \left( \int_0^{\pi/2} x^r \sin(x) \, dx \right) = \int_0^{\pi/2} \frac{d}{dr}(x^r \sin(x)) \, dx = \int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx$$ For the denominator, we need to apply the product rule: $$\frac{d}{dr} \left( r\int_0^{\pi/2} x^r \cos(x) \, dx \right) = \int_0^{\pi/2} x^r \cos(x) \, dx + r\frac{d}{dr} \left( \int_0^{\pi/2} x^r \cos(x) \, dx \right)$$ $$= \int_0^{\pi/2} x^r \cos(x) \, dx + r\int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx$$ Now, we can substitute these derivatives back into the limit: $$L = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{\int_0^{\pi/2} x^r \cos(x) \, dx + r\int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx}$$ At this point, we can see that the limit still appears to be indeterminate. However, if we look closely, we can notice that the second term in the denominator dominates the first term as $r \rightarrow \infty$. Therefore, we can neglect the first term in the denominator and focus on the second term: $$L \approx \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{r\int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx}$$ Now, we can repeat the process of differentiating with respect to $r$. However, since the logarithm function does not grow as fast as the exponential function, it becomes clear that the dominant behavior of the integrals will be determined by the $x^r$ term. To proceed further, we would need to evaluate the integrals, which might not have closed-form expressions in terms of elementary functions. Therefore, we can consider using asymptotic analysis or numerical methods to approximate the limit. In practice, the limit might be evaluated by recognizing that the main contribution to the integrals comes from the region near $x = 0$, where $\sin(x) \approx x$ and $\cos(x) \approx 1 - \frac{x^2}{2}$. This would lead to an approximation of the integrals involving only powers of $x$, which could be integrated explicitly and might provide a clearer path to determining the limit. Without further simplification or approximation, the final answer cannot be determined analytically. If a numerical approach is taken, one would calculate the value of the limit for increasing values of $r$ and observe the convergence pattern. |
0.63662 |
| doubao | Let's use integration by parts. For the numerator, let $u=x^r$ and $dv=\sin(x)dx$. Then $du=rx^{r-1}dx$ and $v=-\cos(x)$. So, \[ \begin{align*} \int_0^{\frac{\pi}{2}}x^r\sin(x)dx&=[-x^r\cos(x)]_0^{\frac{\pi}{2}}+r\int_0^{\frac{\pi}{2}}x^{r-1}\cos(x)dx\\ &=r\int_0^{\frac{\pi}{2}}x^{r-1}\cos(x)dx \end{align*} \] For the denominator, let $u=x^r$ and $dv=\cos(x)dx$. Then $du=rx^{r-1}dx$ and $v=\sin(x)$. So, \[ \begin{align*} r\int_0^{\frac{\pi}{2}}x^r\cos(x)dx&=[r x^r\sin(x)]_0^{\frac{\pi}{2}}-r(r-1)\int_0^{\frac{\pi}{2}}x^{r-2}\sin(x)dx\\ &=r\left(\left(\frac{\pi}{2}\right)^r-r(r-1)\int_0^{\frac{\pi}{2}}x^{r-2}\sin(x)dx\right) \end{align*} \] Dividing the numerator by the denominator and taking the limit as $r\rightarrow\infty$ gives: \[ \lim_{r\rightarrow\infty}\frac{\int_0^{\frac{\pi}{2}}x^r\sin(x)dx}{r\int_0^{\frac{\pi}{2}}x^r\cos(x)dx}=\lim_{r\rightarrow\infty}\frac{r\int_0^{\frac{\pi}{2}}x^{r-1}\cos(x)dx}{r\left(\left(\frac{\pi}{2}\right)^r-r(r-1)\int_0^{\frac{\pi}{2}}x^{r-2}\sin(x)dx\right)}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi} \] |
0.63662 |
| minimax | To solve the given limit problem, we need to evaluate the following expression as $r \rightarrow \infty$: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r\int_0^{\pi/2} x^r \cos(x) \, dx}$$ First, let's consider the behavior of the integrands as $r \rightarrow \infty$ for $x \in (0, \pi/2)$. As $r$ becomes very large, $x^r$ will dominate both integrands, and since $x$ is less than $\pi/2$, $x^r$ will approach zero for any $x$ in the interval $(0, \pi/2)$. However, the rate at which the integrands approach zero will depend on the behavior of $\sin(x)$ and $\cos(x)$. For $x \in (0, \pi/2)$, $\sin(x)$ and $\cos(x)$ are both positive and bounded between 0 and 1. However, $\cos(x)$ has a maximum value of 1 at $x=0$, while $\sin(x)$ starts from 0 at $x=0$ and increases to 1 at $x=\pi/2$. This means that for small $x$, $\sin(x)$ is smaller than $\cos(x)$, but as $x$ approaches $\pi/2$, $\sin(x)$ approaches 1 while $\cos(x)$ approaches 0. Now, let's analyze the integrals separately: 1. $\int_0^{\pi/2} x^r \sin(x) \, dx$: For small $x$, $x^r$ is small and $\sin(x)$ is also small, so the product is very small. As $x$ increases, $\sin(x)$ increases, but $x^r$ decreases rapidly, so the integrand decreases rapidly as well. 2. $\int_0^{\pi/2} x^r \cos(x) \, dx$: For small $x$, $x^r$ is small and $\cos(x)$ is close to 1, so the product is small. As $x$ increases, $\cos(x)$ decreases, and $x^r$ also decreases, so the integrand decreases rapidly as well. As $r \rightarrow \infty$, the integrands will be dominated by the behavior near $x=0$ because the exponential decay of $x^r$ will overshadow the polynomial growth of $\sin(x)$ and $\cos(x)$. Now, let's look at the limit: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r\int_0^{\pi/2} x^r \cos(x) \, dx} = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{\int_0^{\pi/2} x^r \cos(x) \, dx} \cdot \frac{1}{r}$$ As $r \rightarrow \infty$, the term $\frac{1}{r}$ will go to zero, and the integrals themselves will approach zero because the integrands will go to zero for all $x$ in the interval $(0, \pi/2)$. However, the ratio of the integrals will approach a finite limit because the integrands are positive and continuous on the interval, and the ratio of two infinitesimals of the same order is a finite number. To find the exact limit, we can use L'Hôpital's rule, differentiating the numerator and denominator with respect to $r$: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{\int_0^{\pi/2} x^r \cos(x) \, dx} = \lim _{r \rightarrow \infty} \frac{\frac{d}{dr} \int_0^{\pi/2} x^r \sin(x) \, dx}{\frac{d}{dr} \int_0^{\pi/2} x^r \cos(x) \, dx}$$ The derivatives of the integrals with respect to $r$ can be found using Leibniz's rule for differentiating under the integral sign: $$\frac{d}{dr} \int_{a(r)}^{b(r)} f(x, r) \, dx = \int_{a(r)}^{b(r)} \frac{\partial}{\partial r} f(x, r) \, dx + f(b(r), r) \frac{db}{dr} - f(a(r), r) \frac{da}{dr}$$ Applying this rule to our integrals, we get: $$\frac{d}{dr} \int_0^{\pi/2} x^r \sin(x) \, dx = \int_0^{\pi/2} \frac{\partial}{\partial r} (x^r \sin(x)) \, dx = \int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx$$ $$\frac{d}{dr} \int_0^{\pi/2} x^r \cos(x) \, dx = \int_0^{\pi/2} \frac{\partial}{\partial r} (x^r \cos(x)) \, dx = \int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx$$ Now, we can apply L'Hôpital's rule: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{\int_0^{\pi/2} x^r \cos(x) \, dx} = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) \, dx}{\int_0^{\pi/2} x^r \ln(x) \cos(x) \, dx}$$ Since $\ln(x)$ is a smooth function and $\sin(x)$ and $\cos(x)$ are bounded, the integrals will be dominated by the behavior of $x^r \ln(x)$ near $x=0$. As $r \rightarrow \infty$, the integrals will approach zero, but the ratio of the integrals will approach a finite limit because the integrands are positive and continuous on the interval. However, we can see that the integrands $x^r \ln(x) \sin(x)$ and $x^r \ln(x) \cos(x)$ will also approach zero for all $x$ in the interval $(0, \pi/2)$ as $r \rightarrow \infty$. This means that we can apply L'Hôpital's rule repeatedly until we get a non-zero numerator and denominator. The problem is that the derivatives of the integrands with respect to $r$ will still involve $x^r \ln(x)$ terms, and we will not be able to simplify the expression to a form that gives us a clear limit without further information about the behavior of the integrals as $r$ *** ***roceed further, we would need to evaluate the integrals directly or use more advanced techniques to analyze the behavior of the integrands as $r \rightarrow \infty$. This might involve approximations or asymptotic analysis, which is beyond the scope of a simple solution. Therefore, the solution to the given problem is not straightforward and requires more advanced techniques to evaluate the limit. If you have any specific methods or theorems in mind that you would like to apply, please provide further details, and I can assist you accordingly. |
0.63662 |
| kimi | To solve this limit problem, we can use L'Hôpital's Rule, which states that if the limit of the ratio of the derivatives of two functions exists, then the limit of the ratio of the functions themselves also exists and is equal to the limit of the ratio of their derivatives. Given the limit: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx}$$ We can differentiate the numerator and the denominator with respect to \( r \) to apply L'Hôpital's Rule. Let's denote the numerator integral as \( I_1(r) \) and the denominator integral as \( I_2(r) \): $$ I_1(r) = \int_0^{\pi/2} x^r \sin(x) \, dx $$ $$ I_2(r) = \int_0^{\pi/2} x^r \cos(x) \, dx $$ Now, we differentiate \( I_1(r) \) and \( I_2(r) \) with respect to \( r \): $$ \frac{d}{dr} I_1(r) = \int_0^{\pi/2} \ln(x) x^r \sin(x) \, dx $$ $$ \frac{d}{dr} I_2(r) = \int_0^{\pi/2} \ln(x) x^r \cos(x) \, dx $$ Now, applying L'Hôpital's Rule, we get: $$\lim _{r \rightarrow \infty} \frac{\frac{d}{dr} I_1(r)}{\frac{d}{dr} I_2(r)} = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} \ln(x) x^r \sin(x) \, dx}{\int_0^{\pi/2} \ln(x) x^r \cos(x) \, dx}$$ As \( r \) approaches infinity, the integrals will be dominated by the behavior near the upper limit of integration, \( \pi/2 \). However, since \( \ln(x) \) is undefined at \( x = 0 \), we can't directly evaluate the integrals at \( r \to \infty \). Instead, we can consider the behavior of the integrands as \( x \) approaches \( \pi/2 \). As \( x \) approaches \( \pi/2 \), \( \sin(x) \) approaches 1 and \( \cos(x) \) approaches 0. Therefore, the integrand of \( I_1(r) \) approaches \( \ln(\pi/2) \) and the integrand of \( I_2(r) \) approaches 0. Since \( \cos(x) \) goes to 0 faster than \( \sin(x) \), the second integral will be dominated by the term \( \ln(x) \) as \( r \) goes to infinity. Thus, the limit becomes: $$\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} \ln(x) x^r \sin(x) \, dx}{\int_0^{\pi/2} \ln(x) x^r \cos(x) \, dx} = \lim _{r \rightarrow \infty} \frac{\ln(\pi/2)}{0}$$ This limit does not exist because it is of the form \( \frac{c}{0} \), where \( c \) is a constant (in this case, \( \ln(\pi/2) \)). Therefore, the original limit does not exist. |
0.63662 |
| wenxinyiyan | 我们可以利用洛必达法则来求解这个极限。 首先，我们分别对分子和分母求导。 分子的导数为： $\frac{d}{dr} \int_0^{\pi/2} x^r \sin(x) dx = \int_0^{\pi/2} x^r \ln(x) \sin(x) dx$ 分母的导数为： $\frac{d}{dr} r \int_0^{\pi/2} x^r \cos(x) dx = \int_0^{\pi/2} x^r \cos(x) dx + r \int_0^{\pi/2} x^r \ln(x) \cos(x) dx$ 然后，我们计算极限： $\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) dx}{r \int_0^{\pi/2} x^r \cos(x) dx} = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) dx}{\int_0^{\pi/2} x^r \cos(x) dx + r \int_0^{\pi/2} x^r \ln(x) \cos(x) dx}$ 由于 $\ln(x)$ 在 $(0, \pi/2)$ 上是负的，因此当 $r \rightarrow \infty$ 时，分子和分母都趋向于负无穷大。但是，由于 $\sin(x)$ 和 $\cos(x)$ 在 $(0, \pi/2)$ 上都是正的，因此分子和分母的比值趋向于一个常数。 我们可以利用洛必达法则再次计算极限： $\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln(x) \sin(x) dx}{\int_0^{\pi/2} x^r \cos(x) dx + r \int_0^{\pi/2} x^r \ln(x) \cos(x) dx} = \lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \ln^2(x) \sin(x) dx + \int_0^{\pi/2} x^r \ln(x) \cos(x) dx}{\int_0^{\pi/2} x^r \ln(x) \cos(x) dx + \int_0^{\pi/2} x^r \cos(x) dx + r \int_0^{\pi/2} x^r \ln^2(x) \cos(x) dx + r \int_0^{\pi/2} x^r \ln(x) \sin(x) dx}$ 由于 $\ln^2(x)$ 在 $(0, \pi/2)$ 上是正的，因此当 $r \rightarrow \infty$ 时，分子和分母都趋向于正无穷大。但是，由于 $\sin(x)$ 和 $\cos(x)$ 在 $(0, \pi/2)$ 上都是正的，因此分子和分母的比值趋向于一个常数。 我们可以继续利用洛必达法则计算极限，直到分子和分母的比值不再包含 $r$。最终，我们得到： $\lim _{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) dx}{r \int_0^{\pi/2} x^r \cos(x) dx} = \tan(\frac{\pi}{4}) = 1$ |
0.63662 |
| zhipu | To find the limit \(\lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx}\), we need to analyze the behavior of the integrals in the numerator and the denominator as \( r \) approaches infinity. ### Step 1: Behavior of the integrals Consider the integral \( \int_0^{\pi/2} x^r \sin(x) \, dx \). For large \( r \), the term \( x^r \) dominates near \( x = \pi/2 \) because \( x^r \) grows very rapidly as \( x \) approaches \( \pi/2 \). Similarly, consider the integral \( \int_0^{\pi/2} x^r \cos(x) \, dx \). The term \( x^r \) will also dominate this integral for large \( r \). ### Step 2: Approximating the integrals To get a better handle on these integrals, we can approximate them by focusing on the behavior of the integrand near \( x = \pi/2 \). For \( \int_0^{\pi/2} x^r \sin(x) \, dx \): - Near \( x = \pi/2 \), \( \sin(x) \approx 1 \). - Thus, \( \int_0^{\pi/2} x^r \sin(x) \, dx \approx \int_0^{\pi/2} x^r \, dx \). For \( \int_0^{\pi/2} x^r \cos(x) \, dx \): - Near \( x = \pi/2 \), \( \cos(x) \approx 0 \), but more precisely, \( \cos(x) \approx \frac{\pi}{2} - x \) for small \( \epsilon = \frac{\pi}{2} - x \). - Thus, \( \int_0^{\pi/2} x^r \cos(x) \, dx \approx \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx \). ### Step 3: Evaluating the integrals #### Numerator: \[ \int_0^{\pi/2} x^r \, dx = \left[ \frac{x^{r+1}}{r+1} \right]_0^{\pi/2} = \frac{(\pi/2)^{r+1}}{r+1} \] #### Denominator: \[ \int_0^{\pi/2} x^r (\frac{\pi}{2} - x) \, dx = \int_0^{\pi/2} x^r \frac{\pi}{2} \, dx - \int_0^{\pi/2} x^{r+1} \, dx \] \[ = \frac{\pi}{2} \int_0^{\pi/2} x^r \, dx - \int_0^{\pi/2} x^{r+1} \, dx \] \[ = \frac{\pi}{2} \cdot \frac{(\pi/2)^{r+1}}{r+1} - \frac{(\pi/2)^{r+2}}{r+2} \] ### Step 4: Simplifying the expression Now, we need to consider the limit: \[ \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin(x) \, dx}{r \int_0^{\pi/2} x^r \cos(x) \, dx} \approx \lim_{r \rightarrow \infty} \frac{\frac{(\pi/2)^{r+1}}{r+1}}{r \left( \frac{\pi}{2} \cdot \frac{(\pi/2)^{r+1}}{r+1} - \frac{(\pi/2)^{r+2}}{r+2} \right)} \] Simplify the denominator: \[ r \left( \frac{\pi}{2} \cdot \frac{(\pi/2)^{r+1}}{r+1} - \frac{(\pi/2)^{r+2}}{r+2} \right) = r \left( \frac{\pi (\pi/2)^{r+1}}{2(r+1)} - \frac{(\pi/2)^{r+2}}{r+2} \right) \] Factor out \( (\pi/2)^{r+1} \): \[ = r (\pi/2)^{r+1} \left( \frac{\pi}{2(r+1)} - \frac{(\pi/2)}{r+2} \right) \] As \( r \to \infty \), the dominant term in the parentheses is \( \frac{\pi}{2(r+1)} \): \[ \approx r (\pi/2)^{r+1} \cdot \frac{\pi}{2(r+1)} = \frac{r \pi^2}{4(r+1)} (\pi/2)^{r+1} \] ### Step 5: Taking the limit Now, the expression becomes: \[ \lim_{r \rightarrow \infty} \frac{\frac{(\pi/2)^{r+1}}{r+1}}{\frac{r \pi^2}{4(r+1)} (\pi/2)^{r+1}} = \lim_{r \rightarrow \infty} \frac{4}{r \pi^2} \] As \( r \to \infty \), \( \frac{4}{r \pi^2} \to 0 \). ### Conclusion Thus, the limit is: \[ \boxed{0} \] |
0.63662 |
| gpto1 | To find the limit \[ L = \lim_{r \rightarrow \infty} \frac{\int_0^{\pi/2} x^r \sin x \, dx}{r \int_0^{\pi/2} x^r \cos x \, dx}, \] we can use asymptotic analysis for large \( r \). As \( r \rightarrow \infty \), the integrals are dominated by the behavior near the upper limit \( x = \frac{\pi}{2} \) because \( x^r \) becomes significant only when \( x \) is close to \( \frac{\pi}{2} \) (since any number less than \( \frac{\pi}{2} \) raised to a large power becomes negligible). Let's make the substitution \( x = \frac{\pi}{2} - t \), with \( t \geq 0 \) small when \( x \) is near \( \frac{\pi}{2} \). Then \( dx = -dt \), and the limits of integration for \( t \) become from 0 to \( \frac{\pi}{2} \), after changing the sign due to \( dx = -dt \). We approximate the functions near \( x = \frac{\pi}{2} \): \[ x^r \approx \left( \frac{\pi}{2} - t \right)^r = \left( \frac{\pi}{2} \right)^r e^{-r \left( \frac{2t}{\pi} + \frac{2t^2}{\pi^2} \right)} \approx \left( \frac{\pi}{2} \right)^r e^{- \frac{2 r t}{\pi}}. \] \[ \sin x = \sin\left( \frac{\pi}{2} - t \right) \approx \cos t \approx 1 - \frac{t^2}{2}. \] \[ \cos x = \cos\left( \frac{\pi}{2} - t \right) \approx \sin t \approx t - \frac{t^3}{6}. \] Substituting back into the integrals: \[ \int_0^{\pi/2} x^r \sin x \, dx \approx \left( \frac{\pi}{2} \right)^r \int_0^\infty e^{- \frac{2 r t}{\pi}} \left( 1 - \frac{t^2}{2} \right) dt. \] \[ r \int_0^{\pi/2} x^r \cos x \, dx \approx r \left( \frac{\pi}{2} \right)^r \int_0^\infty e^{- \frac{2 r t}{\pi}} \left( t - \frac{t^3}{6} \right) dt. \] Evaluating the integrals using standard Laplace transforms (noting that \( e^{-a t} t^n \) integrates to \( \frac{n!}{a^{n+1}} \)): \[ \int_0^\infty e^{-a t} dt = \frac{1}{a}, \quad \int_0^\infty e^{-a t} t^n dt = \frac{n!}{a^{n+1}}. \] With \( a = \frac{2 r}{\pi} \), we find: \[ \int_0^\infty e^{- \frac{2 r t}{\pi}} dt = \frac{\pi}{2 r}, \quad \int_0^\infty e^{- \frac{2 r t}{\pi}} t^2 dt = \frac{2! \pi^3}{(2 r)^3}. \] Substituting back: \[ \int_0^{\pi/2} x^r \sin x \, dx \approx \left( \frac{\pi}{2} \right)^r \left( \frac{\pi}{2 r} - \frac{\pi^3}{(2 r)^3} \right). \] \[ r \int_0^{\pi/2} x^r \cos x \, dx \approx r \left( \frac{\pi}{2} \right)^r \left( \frac{\pi^2}{(2 r)^2} - \frac{\pi^4}{(2 r)^4} \right). \] Simplifying the ratio \( L \): \[ L = \lim_{r \rightarrow \infty} \frac{\left( \frac{\pi}{2 r} - \frac{\pi^3}{(2 r)^3} \right)}{\left( \frac{\pi^2}{(2 r)^2} - \frac{\pi^4}{(2 r)^4} \right)} = \lim_{r \rightarrow \infty} \frac{\frac{\pi}{2 r}}{\frac{\pi^2}{(2 r)^2}} = \frac{1}{\pi}, \] since the higher-order terms become negligible as \( r \) approaches infinity. **Answer:** **\(\displaystyle \frac{1}{\pi}\)** |
0.63662 |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-47d5cfb8-cc96-9586-9791-b3b0a1b4a8d7', 'request_id': '47d5cfb8-cc96-9586-9791-b3b0a1b4a8d7'} | 0.63662 |