| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given problem, we will first simplify the expression for \( g(x) \) and then find the limit of \( g(x)/x \) as \( x \) goes to infinity. Given: \[ g(x) = \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r} \] Let's simplify the expression inside the limit: \[ g(x) = \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r} \] We can rewrite the expression using the binomial expansion: \[ (x+1)^{r+1} = (x+1)\left(1 + r\ln(x+1) + O(r^2)\right) \] \[ x^{r+1} = x\left(1 + r\ln(x) + O(r^2)\right) \] Thus: \[ (x+1)^{r+1} - x^{r+1} = (x+1)\left(1 + r\ln(x+1) + O(r^2)\right) - x\left(1 + r\ln(x) + O(r^2)\right) \] \[ = (x+1) + r(x+1)\ln(x+1) + O(r^2) - x - rx\ln(x) + O(r^2) \] \[ = 1 + r\left((x+1)\ln(x+1) - x\ln(x)\right) + O(r^2) \] Taking the limit as \( r \to 0 \): \[ g(x) = \lim_{r \to 0} \left(1 + r\left((x+1)\ln(x+1) - x\ln(x)\right) + O(r^2)\right)^{1/r} \] Using the fact that \( \lim_{r \to 0} (1 + ry)^{1/r} = e^y \): \[ g(x) = e^{(x+1)\ln(x+1) - x\ln(x)} \] \[ = e^{\ln(x+1)^{x+1} - \ln(x)^x} \] \[ = e^{\ln\left(\frac{(x+1)^{x+1}}{x^x}\right)} \] \[ = \frac{(x+1)^{x+1}}{x^x} \] Now, we need to find the limit of \( g(x)/x \) as \( x \) goes to infinity: \[ \lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{(x+1)^{x+1}}{x^{x+1}} \] \[ = \lim_{x \to \infty} \left(\frac{x+1}{x}\right)^{x+1} \] \[ = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x+1} \] Using the fact that \( \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e \): \[ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x+1} = e \] Therefore, the limit of \( g(x)/x \) as \( x \) goes to infinity is: \[ \boxed{e} \] |
2.7182818 |
| deepseekr1 | To find the limit of \( g(x)/x \) as \( x \) approaches infinity, where \( g(x) = \lim_{r \rightarrow 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \), we start by analyzing the inner limit. First, we rewrite the expression inside the limit: \[ A(r) = (x+1)^{r+1} - x^{r+1} = x^{r+1} \left( \left(1 + \frac{1}{x}\right)^{r+1} - 1 \right) \] For small \( r \), we approximate \( \left(1 + \frac{1}{x}\right)^{r+1} \) using a Taylor expansion: \[ \left(1 + \frac{1}{x}\right)^{r+1} \approx 1 + (r+1) \ln\left(1 + \frac{1}{x}\right) \] Thus, \[ A(r) \approx x^{r+1} \left( (r+1) \ln\left(1 + \frac{1}{x}\right) \right) \] Factoring out \( x^{r+1} \), we get: \[ A(r) \approx x^{r+1} \cdot \frac{1}{x} \left( 1 + r x \ln\left(1 + \frac{1}{x}\right) \right) = x^r \left( 1 + r x \ln\left(1 + \frac{1}{x}\right) \right) \] Taking the logarithm of \( A(r) \): \[ \ln A(r) \approx r \ln x + \ln\left(1 + r x \ln\left(1 + \frac{1}{x}\right)\right) \] For small \( r \), we approximate the logarithm: \[ \ln\left(1 + r x \ln\left(1 + \frac{1}{x}\right)\right) \approx r x \ln\left(1 + \frac{1}{x}\right) \] Thus, \[ \ln A(r) \approx r \ln x + r x \ln\left(1 + \frac{1}{x}\right) \] Dividing by \( r \) and taking the limit as \( r \to 0 \): \[ \ln L = \lim_{r \to 0} \frac{\ln A(r)}{r} = \ln x + x \ln\left(1 + \frac{1}{x}\right) \] Exponentiating this result gives: \[ L = x \exp\left( x \ln\left(1 + \frac{1}{x}\right) \right) = x \left(1 + \frac{1}{x}\right)^x \] Thus, \( g(x) = x \left(1 + \frac{1}{x}\right)^x \). Therefore, \( \frac{g(x)}{x} = \left(1 + \frac{1}{x}\right)^x \), and as \( x \) approaches infinity, this limit is \( e \). The final answer is: \[ \boxed{e} \] |
2.7182818 |
| gpto1 | To solve this problem, we need to find \( \lim_{x \to \infty} \frac{g(x)}{x} \), where \[ g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{\frac{1}{r}}. \] **Step 1: Evaluate \( g(x) \)** First, we analyze the expression inside the limit as \( r \to 0 \). We can rewrite \( (x+1)^{r+1} \) and \( x^{r+1} \) as: \[ (x+1)^{r+1} = (x+1)^{1 + r} = (x+1) \cdot (x+1)^r, \] \[ x^{r+1} = x^{1 + r} = x \cdot x^r. \] Thus, \[ (x+1)^{r+1} - x^{r+1} = (x+1) (x+1)^r - x x^r. \] Since \( r \) is approaching 0, we can expand \( (x+1)^r \) and \( x^r \) using the exponential function: \[ (x+1)^r = e^{r \ln(x+1)} \approx 1 + r \ln(x+1), \] \[ x^r = e^{r \ln x} \approx 1 + r \ln x. \] Now, the expression becomes: \[ (x+1) [1 + r \ln(x+1)] - x [1 + r \ln x] = (x+1)(1 + r \ln(x+1)) - x(1 + r \ln x). \] Simplify this expression: \[ [x + 1 + r(x+1)\ln(x+1)] - [x + r x \ln x] = 1 + r[(x+1)\ln(x+1) - x \ln x]. \] Thus, the original expression becomes: \[ \left(1 + r[(x+1)\ln(x+1) - x \ln x]\right)^{\frac{1}{r}}. \] **Step 2: Simplify the Limit** As \( r \to 0 \), we can use the approximation \( \lim_{r \to 0} (1 + r a)^{\frac{1}{r}} = e^a \) for any real number \( a \). Applying this, we have: \[ g(x) = e^{(x+1)\ln(x+1) - x \ln x}. \] **Step 3: Simplify the Exponent** Compute the exponent: \[ E = (x+1)\ln(x+1) - x \ln x. \] Expand \( (x+1)\ln(x+1) \) as: \[ (x+1)\ln(x+1) = x \ln(x+1) + \ln(x+1). \] Also, note that \( \ln(x+1) = \ln x + \ln\left(1 + \frac{1}{x}\right) \approx \ln x + \frac{1}{x} - \frac{1}{2 x^2} \) for large \( x \). Substituting back, we get: \[ E = x[\ln x + \frac{1}{x} - \frac{1}{2 x^2}] + [\ln x + \frac{1}{x} - \frac{1}{2 x^2}] - x \ln x. \] Simplify: \[ E = x \left( \frac{1}{x} - \frac{1}{2 x^2} \right) + \ln x + \frac{1}{x} - \frac{1}{2 x^2} - x \ln x + x \ln x. \] This simplifies to: \[ E = 1 - \frac{1}{2 x} + \ln x + \frac{1}{x} - \frac{1}{2 x^2}. \] Ignoring terms that vanish as \( x \to \infty \), the exponent \( E \) approximates to: \[ E \approx \ln x + 1. \] **Step 4: Compute \( g(x) \)** Now, \( g(x) \) becomes: \[ g(x) = e^{\ln x + 1} = x e. \] **Step 5: Find the Limit** Finally, compute the desired limit: \[ \lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{x e}{x} = e. \] **Answer:** \[ e \] |
2.7182818 |
| GPT-4o | To solve the problem, we start by analyzing the function \( g(x) \). Given that \( g(x) = \lim_{r \to 0} ((x+1)^{r+1} - x^{r+1})^{1/r} \), we first need to simplify and understand the expression inside the limit. Consider the expression \( h(r) = ((x+1)^{r+1} - x^{r+1})^{1/r} \). To proceed, let's expand \( (x+1)^{r+1} \) and \( x^{r+1} \) using the binomial theorem for small \( r \): \[ (x+1)^{r+1} = e^{(r+1) \ln(x+1)} \approx e^{r \ln(x+1) + \ln(x+1)} = (x+1)e^{r \ln(x+1)} \] \[ x^{r+1} = e^{(r+1) \ln x} \approx e^{r \ln x + \ln x} = x e^{r \ln x} \] Now, let's consider the difference: \[ (x+1)^{r+1} - x^{r+1} \approx (x+1)e^{r \ln(x+1)} - x e^{r \ln x} \] For small values of \( r \), we can approximate \( e^{r \ln(x+1)} \approx 1 + r \ln(x+1) \) and \( e^{r \ln x} \approx 1 + r \ln x \). Thus, \[ (x+1) (1 + r \ln(x+1)) - x (1 + r \ln x) \approx (x+1)(1 + r \ln(x+1)) - x(1 + r \ln x) \] \[ = (x+1 + r (x+1) \ln(x+1)) - (x + r x \ln x) = 1 + r ((x+1) \ln(x+1) - x \ln x) \] Now, let's consider taking the limit as \( r \to 0 \): \[ ((x+1)^{r+1} - x^{r+1})^{1/r} \approx (1 + r ((x+1) \ln(x+1) - x \ln x))^{1/r} \] Using the approximation \((1 + ar)^{1/r} \approx e^a\) for small \( r \), we get: \[ ((x+1)^{r+1} - x^{r+1})^{1/r} \approx e^{(x+1) \ln(x+1) - x \ln x} \] We now need to find the limit of \( g(x)/x \) as \( x \to \infty \): \[ g(x) = e^{(x+1) \ln(x+1) - x \ln x} \] To find \( \lim_{x \to \infty} \frac{g(x)}{x} \): \[ \frac{g(x)}{x} = \frac{e^{(x+1) \ln(x+1) - x \ln x}}{x} \] For large \( x \): \[ (x+1) \ln(x+1) \approx x \ln x + \ln x + 1 \] \[ (x+1) \ln(x+1) - x \ln x \approx x \ln x + \ln x + 1 - x \ln x = \ln x + 1 \] We then have: \[ g(x) \approx e^{\ln x + 1} = e^{\ln x} \cdot e = x \cdot e \] Therefore, \[ \frac{g(x)}{x} \approx \frac{x \cdot e}{x} = e \] Thus, the limit is: \[ \boxed{e} \] |
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| llama3.1-70B-Instruct | ## Step 1: Understand the given function g(x) and its limit The function $g(x)$ is defined as the limit of $((x+1)^{r+1}-x^{r+1})^{1/r}$ as $r$ approaches 0. ## Step 2: Analyze the expression inside the limit We can rewrite the expression inside the limit as $((x+1)^{r+1}-x^{r+1})^{1/r} = (x+1) \cdot ((1 + 1/x)^{r+1} - 1)^{1/r}$. ## Step 3: Use the binomial expansion for the expression inside the limit Using the binomial expansion, we can expand $(1 + 1/x)^{r+1}$ as $1 + (r+1)/x + O(1/x^2)$. ## Step 4: Simplify the expression inside the limit Substituting the binomial expansion into the expression, we get $((x+1)^{r+1}-x^{r+1})^{1/r} = (x+1) \cdot ((1 + (r+1)/x + O(1/x^2)) - 1)^{1/r}$. ## Step 5: Take the limit as r approaches 0 Taking the limit as $r$ approaches 0, we get $g(x) = (x+1) \cdot \lim_{r \to 0} ((r+1)/x + O(1/x^2))^{1/r}$. ## Step 6: Evaluate the limit Evaluating the limit, we get $g(x) = (x+1) \cdot (1/x)$. ## Step 7: Find the limit of g(x)/x as x goes to infinity Now, we need to find the limit of $g(x)/x$ as $x$ goes to infinity. Substituting $g(x) = (x+1) \cdot (1/x)$, we get $\lim_{x \to \infty} g(x)/x = \lim_{x \to \infty} ((x+1) \cdot (1/x))/x$. ## Step 8: Simplify the expression Simplifying the expression, we get $\lim_{x \to \infty} g(x)/x = \lim_{x \to \infty} (x+1)/x^2$. ## Step 9: Evaluate the limit Evaluating the limit, we get $\lim_{x \to \infty} g(x)/x = 1$. The final answer is: $\boxed{1}$ |
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| llama3.1-405B-Instruct | To find the limit of $g(x)/x$ as $x$ goes to infinity, we can start by analyzing the expression for $g(x)$. ## Step 1: Rewrite the expression for $g(x)$ We can rewrite the expression for $g(x)$ as follows: $$g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r}$$ ## Step 2: Expand the expression using the binomial theorem Using the binomial theorem, we can expand the expression for $g(x)$ as follows: $$g(x) = \lim_{r \to 0} \left( x^{r+1} + (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots - x^{r+1} \right)^{1/r}$$ ## Step 3: Simplify the expression Simplifying the expression, we get: $$g(x) = \lim_{r \to 0} \left( (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots \right)^{1/r}$$ ## Step 4: Take the limit as $r$ approaches 0 Taking the limit as $r$ approaches 0, we get: $$g(x) = \lim_{r \to 0} \left( (r+1)x^r \right)^{1/r}$$ ## Step 5: Simplify the expression Simplifying the expression, we get: $$g(x) = x \lim_{r \to 0} (r+1)^{1/r}$$ ## Step 6: Evaluate the limit Evaluating the limit, we get: $$g(x) = x \cdot e$$ ## Step 7: Find the limit of $g(x)/x$ as $x$ goes to infinity Finally, we can find the limit of $g(x)/x$ as $x$ goes to infinity: $$\lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{x \cdot e}{x} = e$$ The final answer is: $\boxed{e}$ |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to find the limit of \( \frac{g(x)}{x} \) as \( x \) approaches infinity, where \( g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \). First, let's analyze the expression inside the limit: \[ g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \] We can rewrite the expression inside the limit using the binomial expansion for small \( r \): \[ (x+1)^{r+1} = (x+1) \cdot (x+1)^r \] \[ x^{r+1} = x \cdot x^r \] For small \( r \), we can approximate \( (x+1)^r \) and \( x^r \) using the first term of the binomial expansion: \[ (x+1)^r \approx 1 + r \ln(x+1) \] \[ x^r \approx 1 + r \ln(x) \] Substituting these approximations back into the expression, we get: \[ (x+1)^{r+1} \approx (x+1)(1 + r \ln(x+1)) = x + 1 + r(x+1) \ln(x+1) \] \[ x^{r+1} \approx x(1 + r \ln(x)) = x + r x \ln(x) \] Thus, \[ (x+1)^{r+1} - x^{r+1} \approx (x + 1 + r(x+1) \ln(x+1)) - (x + r x \ln(x)) \] \[ = 1 + r(x+1) \ln(x+1) - r x \ln(x) \] \[ = 1 + r \left( (x+1) \ln(x+1) - x \ln(x) \right) \] Now, we need to take the limit as \( r \to 0 \): \[ g(x) = \lim_{r \to 0} \left( 1 + r \left( (x+1) \ln(x+1) - x \ln(x) \right) \right)^{1/r} \] For small \( r \), the expression inside the limit can be approximated by: \[ \left( 1 + r \left( (x+1) \ln(x+1) - x \ln(x) \right) \right)^{1/r} \approx e^{\frac{(x+1) \ln(x+1) - x \ln(x)}{1}} \] \[ = e^{(x+1) \ln(x+1) - x \ln(x)} \] Simplifying the exponent: \[ (x+1) \ln(x+1) - x \ln(x) = (x+1) \ln(x+1) - x \ln(x) \] \[ = x \ln(x+1) + \ln(x+1) - x \ln(x) \] \[ = x (\ln(x+1) - \ln(x)) + \ln(x+1) \] \[ = x \ln \left( \frac{x+1}{x} \right) + \ln(x+1) \] \[ = x \ln \left( 1 + \frac{1}{x} \right) + \ln(x+1) \] As \( x \to \infty \), \( \ln \left( 1 + \frac{1}{x} \right) \to 0 \) and \( \ln(x+1) \approx \ln(x) \): \[ g(x) \approx e^{x \cdot 0 + \ln(x)} = e^{\ln(x)} = x \] Therefore, \[ \frac{g(x)}{x} \approx \frac{x}{x} = 1 \] Thus, the limit of \( \frac{g(x)}{x} \) as \( x \) goes to infinity is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | To solve the given problem, we will first simplify the expression for $g(x)$ and then find the limit of $g(x)/x$ as $x$ goes to infinity. Given: $$g(x) = \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r}$$ Let's simplify the expression inside the limit: \begin{align*} g(x) &= \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r} \\ &= \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r} \\ &= \lim_{r \to 0} \left((x+1)^{r}(x+1) - x^{r}x\right)^{1/r} \\ &= \lim_{r \to 0} \left(x(x+1)^{r} + (x+1)^{r} - x^{r}x\right)^{1/r} \\ &= \lim_{r \to 0} \left(x\left((x+1)^{r} - x^{r}\right) + (x+1)^{r}\right)^{1/r} \end{align*} Since $\lim_{r \to 0} x^{r} = 1$ for any $x > 0$, we can simplify further: \begin{align*} g(x) &= \lim_{r \to 0} \left(x\left((x+1)^{r} - 1\right) + (x+1)^{r}\right)^{1/r} \\ \end{align*} Now, we need to find the limit of $g(x)/x$ as $x$ goes to infinity: \begin{align*} \lim_{x \to \infty} \frac{g(x)}{x} &= \lim_{x \to \infty} \frac{\lim_{r \to 0} \left(x\left((x+1)^{r} - 1\right) + (x+1)^{r}\right)^{1/r}}{x} \\ &= \lim_{x \to \infty} \lim_{r \to 0} \left(\left((x+1)^{r} - 1\right) + \frac{(x+1)^{r}}{x}\right)^{1/r} \end{align*} As $x \to \infty$, $\frac{(x+1)^{r}}{x} \to 0$, and we are left with: \begin{align*} \lim_{x \to \infty} \frac{g(x)}{x} &= \lim_{r \to 0} \left((x+1)^{r} - 1\right)^{1/r} \end{align*} Now, we can use the fact that $\lim_{r \to 0} (1 + y)^{1/r} = e^{y}$ for any $y \in \mathbb{R}$: \begin{align*} \lim_{x \to \infty} \frac{g(x)}{x} &= \lim_{r \to 0} \left((x+1)^{r} - 1\right)^{1/r} \\ &= \lim_{r \to 0} \left(1 + r\ln(x+1)\right)^{1/r} \\ &= e^{\ln(x+1)} \\ &= x + 1 \end{align*} However, since we are taking the limit as $x \to \infty$, we find: \begin{align*} \lim_{x \to \infty} \frac{g(x)}{x} &= \lim_{x \to \infty} (x + 1) \\ &= \infty \end{align*} Therefore, the limit of $g(x)/x$ as $x$ goes to infinity is $\boxed{\infty}$. However, this result is not correct. The correct solution is as follows: Given: $$g(x) = \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r}$$ We can rewrite the expression inside the limit using the binomial theorem: \begin{align*} g(x) &= \lim_{r \to 0} \left((x+1)^{r+1} - x^{r+1}\right)^{1/r} \\ &= \lim_{r \to 0} \left((x+1)^{r}(x+1) - x^{r}x\right)^{1/r} \\ &= \lim_{r \to 0} \left(x(x+1)^{r} + (x+1)^{r} - x^{r}x\right)^{1/r} \\ &= \lim_{r \to 0} \left(x\left((x+1)^{r} - x^{r}\right) + (x+1)^{r}\right)^{1/r} \\ &= \lim_{r \to 0} \left(x\left(\binom{r}{0}x^{r} + \binom{r}{1}x^{r-1} + \binom{r}{2}x^{r-2} + \cdots + \binom{r}{r} - x^{r}\right) + (x+1)^{r}\right)^{1/r} \\ &= \lim_{r \to 0} \left(x\left(1 + rx^{r-1} + \binom{r}{2}x^{r-2} + \cdots + 1 - x^{r}\right) + (x+1)^{r}\right)^{1/r} \end{align*} As $r \to 0$, all terms in the parentheses except for $1$ and $rx^{r-1}$ will go to $0$, and we are left with: \begin{align*} g(x) &= \lim_{r \to 0} \left(x\left(1 + rx^{r-1}\right) + (x+1)^{r}\right)^{1/r} \\ &= \lim_{r \to 0} \left(x + rx^{r}\right)^{1/r} \\ &= \lim_{r \to 0} x^{1/r}\left(1 + r\right)^{1/r} \end{align*} We can use the fact that $\lim_{r \to 0} (1 + r)^{1/r} = e$: \begin{align*} g(x) &= \lim_{r \to 0} x^{1/r}e \\ &= e\lim_{r \to 0} x^{1/r} \end{align*} As $r \to 0$, $x^{1/r} \to \infty$ for any $x > 0$, and we find: \begin{align*} g(x) &= e\lim_{r \to 0} x^{1/r} \\ &= e\cdot\infty \\ &= \infty \end{align*} Therefore, the limit of $g(x)/x$ as $x$ goes to infinity is $\boxed{e}$. |
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| baichuan4 | To find the limit of $g(x)/x$ as $x$ goes to infinity, we first need to understand the behavior of $g(x)$ as defined by the limit: $$g(x)=\lim _{r \rightarrow 0}((x+1)^{r+1}-x^{r+1})^{1/r}$$ Let's simplify this expression by factoring out $x^{r+1}$: $$g(x)=\lim _{r \rightarrow 0} \left(x^{r+1}\left((1 + \frac{1}{x})^{r+1} - 1\right)\right)^{1/r}$$ Now, we can use the known limit: $$\lim_{r \rightarrow 0} (1 + r)^{\frac{1}{r}} = e$$ to identify that: $$\lim_{r \rightarrow 0} (1 + \frac{1}{x})^{r+1} = (1 + \frac{1}{x})^1 = 1 + \frac{1}{x}$$ when $r$ approaches 0. Therefore, we can rewrite $g(x)$ as: $$g(x)=\lim _{r \rightarrow 0} \left(x^{r+1}(1 + \frac{1}{x} - 1)\right)^{1/r} = \lim _{r \rightarrow 0} \left(x^{r+1} \cdot \frac{1}{x}\right)^{1/r}$$ Simplifying further: $$g(x)=\lim _{r \rightarrow 0} \left(x^r\right)^{1/r} = \lim _{r \rightarrow 0} x^{1 + \frac{1}{r} - \frac{1}{r}} = x$$ So, $g(x) = x$. Now, we can find the limit of $g(x)/x$ as $x$ goes to infinity: $$\lim_{x \rightarrow \infty} \frac{g(x)}{x} = \lim_{x \rightarrow \infty} \frac{x}{x} = 1$$ Therefore, the limit of $g(x)/x$ as $x$ goes to infinity is $1$. |
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| doubao | Let's first find $g(x)$. \[ \begin{align*} g(x)&=\lim_{r\rightarrow 0}((x+1)^{r+1}-x^{r+1})^{1/r}\\ &=\lim_{r\rightarrow 0}e^{\frac{1}{r}\ln((x+1)^{r+1}-x^{r+1})}\\ &=\lim_{r\rightarrow 0}e^{\frac{r+1}{r}\ln(x+1)-\frac{r+1}{r}\ln x}\\ &=\lim_{r\rightarrow 0}e^{(1+\frac{1}{r})\ln(x+1)-(1+\frac{1}{r})\ln x}\\ &=e^{\lim_{r\rightarrow 0}[(1+\frac{1}{r})\ln(x+1)-(1+\frac{1}{r})\ln x]}\\ &=e^{\ln(x+1)-\ln x}\\ &=e^{\ln\frac{x+1}{x}}\\ &=\frac{x+1}{x} \end{align*} \] Now we find the limit of $\frac{g(x)}{x}$ as $x$ goes to infinity: \[ \lim_{x\rightarrow\infty}\frac{g(x)}{x}=\lim_{x\rightarrow\infty}\frac{\frac{x+1}{x}}{x}=\lim_{x\rightarrow\infty}\frac{x+1}{x^2}=0. \] |
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| minimax | To find the limit of $g(x)/x$ as $x$ goes to infinity, we first need to evaluate $g(x)$ for every positive real number $x$. We are given that $$g(x)=\lim _{r \rightarrow 0}((x+1)^{r+1}-x^{r+1})^{1/r}.$$ We can use the fact that for small values of $r$, $(1 + r)^{1/r}$ approaches $e$ (where $e$ is the base of the natural logarithm). However, we need to be careful with the expression inside the limit, as it is not exactly in the form of $(1 + r)^{1/r}$. Let's rewrite the expression inside the limit by factoring out $x^{r+1}$: $$((x+1)^{r+1}-x^{r+1})^{1/r} = (x^{r+1}((1+\frac{1}{x})^{r+1}-1))^{1/r}.$$ Now, we can apply the limit: $$g(x) = \lim_{r \rightarrow 0} (x^{r+1}((1+\frac{1}{x})^{r+1}-1))^{1/r}.$$ We can separate the limit into two parts: $$g(x) = \lim_{r \rightarrow 0} x^{(r+1)/r} \cdot \lim_{r \rightarrow 0} ((1+\frac{1}{x})^{r+1}-1)^{1/r}.$$ The first limit is straightforward: $$\lim_{r \rightarrow 0} x^{(r+1)/r} = x^{\lim_{r \rightarrow 0} (1+1/r)} = x.$$ For the second limit, we can apply the fact that $(1+\frac{1}{x})^{r+1}-1$ is close to $\frac{r+1}{x}$ when $r$ is small, so we can approximate it as: $$\lim_{r \rightarrow 0} ((1+\frac{1}{x})^{r+1}-1)^{1/r} \approx \lim_{r \rightarrow 0} \left(\frac{r+1}{x}\right)^{1/r}.$$ Now, we can use the fact that $\lim_{r \rightarrow 0} (1 + r)^{1/r} = e$ to rewrite the expression as: $$\lim_{r \rightarrow 0} \left(\frac{r+1}{x}\right)^{1/r} = \lim_{r \rightarrow 0} \left(\frac{1}{x}\right)^{1/r} \cdot \lim_{r \rightarrow 0} (1 + r)^{1/r} = \frac{1}{x^{1/r}} \cdot e.$$ Since $r$ approaches 0, $x^{1/r}$ approaches infinity, and thus $1/x^{1/r}$ approaches 0. Therefore, the second limit is 0. So, we have: $$g(x) = x \cdot 0 = 0.$$ Now, we want to find the limit of $g(x)/x$ as $x$ goes to infinity: $$\lim_{x \rightarrow \infty} \frac{g(x)}{x} = \lim_{x \rightarrow \infty} \frac{0}{x} = 0.$$ Therefore, the limit of $g(x)/x$ as $x$ goes to infinity is 0. |
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| kimi | To solve the problem, we first need to understand the function \( g(x) \). The function \( g(x) \) is defined as: \[ g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \] Let's simplify the expression inside the limit: \[ (x+1)^{r+1} - x^{r+1} \] Using the binomial theorem, we can expand \( (x+1)^{r+1} \) as: \[ (x+1)^{r+1} = \sum_{k=0}^{r+1} \binom{r+1}{k} x^k \] The first few terms of this expansion are: \[ (x+1)^{r+1} = x^{r+1} + (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots + 1 \] So, \[ (x+1)^{r+1} - x^{r+1} = (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots + 1 \] Now, let's consider the limit as \( r \to 0 \). For small \( r \), the dominant term in the expression above is \( (r+1)x^r \). The other terms will approach 0 as \( r \to 0 \). Therefore, we can approximate: \[ (x+1)^{r+1} - x^{r+1} \approx (r+1)x^r \] Now, we can take the limit: \[ g(x) = \lim_{r \to 0} \left( (r+1)x^r \right)^{1/r} \] \[ g(x) = \lim_{r \to 0} (r+1)^{1/r} x \] For the limit to exist, \( (r+1)^{1/r} \) must approach a finite value as \( r \to 0 \). The limit of \( (r+1)^{1/r} \) as \( r \to 0 \) is \( e \), the base of the natural logarithm. So: \[ g(x) = e \cdot x \] Now, we want to find the limit of \( g(x)/x \) as \( x \) goes to infinity: \[ \lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{e \cdot x}{x} \] \[ \lim_{x \to \infty} \frac{g(x)}{x} = e \cdot \lim_{x \to \infty} \frac{x}{x} \] \[ \lim_{x \to \infty} \frac{g(x)}{x} = e \cdot 1 \] \[ \lim_{x \to \infty} \frac{g(x)}{x} = e \] So, the limit of \( g(x)/x \) as \( x \) goes to infinity is \( e \). |
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| wenxinyiyan | 首先，我们将$g(x)$的表达式进行变形，得到： $g(x) = \lim_{r \rightarrow 0} ((x+1)^{r+1} - x^{r+1})^{1/r}$ $= \lim_{r \rightarrow 0} ((x+1)^{r+1} - x^{r+1})^{\frac{1}{r+1} \cdot \frac{r+1}{r}}$ $= \lim_{r \rightarrow 0} ((x+1) - x)^{\frac{1}{r+1}}$ $= \lim_{r \rightarrow 0} (1)^{\frac{1}{r+1}}$ $= 1$ 所以，$g(x) = 1$。 接下来，我们计算$\lim_{x \rightarrow \infty} \frac{g(x)}{x}$： $\lim_{x \rightarrow \infty} \frac{g(x)}{x} = \lim_{x \rightarrow \infty} \frac{1}{x} = 0$ 所以，$\lim_{x \rightarrow \infty} \frac{g(x)}{x} = 0$。 |
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| zhipu | To find the limit of \( \frac{g(x)}{x} \) as \( x \) goes to infinity, we first need to understand the function \( g(x) \). Given: \[ g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \] Let's analyze the expression inside the limit. We can rewrite it as: \[ (x+1)^{r+1} - x^{r+1} = (x+1) \cdot (x+1)^r - x \cdot x^r \] Now, let's consider the limit as \( r \to 0 \). We know that \( a^r \to 1 \) for any positive \( a \) as \( r \to 0 \). Therefore: \[ (x+1)^r \to 1 \quad \text{and} \quad x^r \to 1 \] So, the expression inside the limit becomes: \[ (x+1) \cdot 1 - x \cdot 1 = x + 1 - x = 1 \] However, this simplification might be too naive. We need to be more careful with the limit process. Let's use the logarithm to simplify the expression: \[ \ln \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} = \frac{1}{r} \ln \left( (x+1)^{r+1} - x^{r+1} \right) \] Now, let's expand \( (x+1)^{r+1} \) and \( x^{r+1} \) using the binomial series for small \( r \): \[ (x+1)^{r+1} \approx x^{r+1} + (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots \] \[ x^{r+1} = x^{r+1} \] Subtracting these, we get: \[ (x+1)^{r+1} - x^{r+1} \approx (r+1)x^r + \frac{(r+1)r}{2}x^{r-1} + \cdots \] For small \( r \), the dominant term is \( (r+1)x^r \). Therefore: \[ \ln \left( (x+1)^{r+1} - x^{r+1} \right) \approx \ln \left( (r+1)x^r \right) = \ln(r+1) + r \ln x \] Thus: \[ \frac{1}{r} \ln \left( (x+1)^{r+1} - x^{r+1} \right) \approx \frac{1}{r} \left( \ln(r+1) + r \ln x \right) \] As \( r \to 0 \), \( \ln(r+1) \approx r \) (using the approximation \( \ln(1+y) \approx y \) for small \( y \)). So: \[ \frac{1}{r} \left( \ln(r+1) + r \ln x \right) \approx \frac{1}{r} \left( r + r \ln x \right) = 1 + \ln x \] Therefore: \[ \ln g(x) \approx 1 + \ln x \] Exponentiating both sides, we get: \[ g(x) \approx e^{1 + \ln x} = e \cdot e^{\ln x} = e \cdot x \] Thus, \( g(x) \approx ex \) for large \( x \). Now, we need to find the limit of \( \frac{g(x)}{x} \) as \( x \to \infty \): \[ \lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{ex}{x} = e \] So, the limit of \( \frac{g(x)}{x} \) as \( x \) goes to infinity is: \[ \boxed{e} \] |
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| gpto1 | To solve this problem, we need to find \( \lim_{x \to \infty} \frac{g(x)}{x} \), where \[ g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{\frac{1}{r}}. \] **Step 1: Evaluate \( g(x) \)** First, we analyze the expression inside the limit as \( r \to 0 \). We can rewrite \( (x+1)^{r+1} \) and \( x^{r+1} \) as: \[ (x+1)^{r+1} = (x+1)^{1 + r} = (x+1) \cdot (x+1)^r, \] \[ x^{r+1} = x^{1 + r} = x \cdot x^r. \] Thus, \[ (x+1)^{r+1} - x^{r+1} = (x+1) (x+1)^r - x x^r. \] Since \( r \) is approaching 0, we can expand \( (x+1)^r \) and \( x^r \) using the exponential function: \[ (x+1)^r = e^{r \ln(x+1)} \approx 1 + r \ln(x+1), \] \[ x^r = e^{r \ln x} \approx 1 + r \ln x. \] Now, the expression becomes: \[ (x+1) [1 + r \ln(x+1)] - x [1 + r \ln x] = (x+1)(1 + r \ln(x+1)) - x(1 + r \ln x). \] Simplify this expression: \[ [x + 1 + r(x+1)\ln(x+1)] - [x + r x \ln x] = 1 + r[(x+1)\ln(x+1) - x \ln x]. \] Thus, the original expression becomes: \[ \left(1 + r[(x+1)\ln(x+1) - x \ln x]\right)^{\frac{1}{r}}. \] **Step 2: Simplify the Limit** As \( r \to 0 \), we can use the approximation \( \lim_{r \to 0} (1 + r a)^{\frac{1}{r}} = e^a \) for any real number \( a \). Applying this, we have: \[ g(x) = e^{(x+1)\ln(x+1) - x \ln x}. \] **Step 3: Simplify the Exponent** Compute the exponent: \[ E = (x+1)\ln(x+1) - x \ln x. \] Expand \( (x+1)\ln(x+1) \) as: \[ (x+1)\ln(x+1) = x \ln(x+1) + \ln(x+1). \] Also, note that \( \ln(x+1) = \ln x + \ln\left(1 + \frac{1}{x}\right) \approx \ln x + \frac{1}{x} - \frac{1}{2 x^2} \) for large \( x \). Substituting back, we get: \[ E = x[\ln x + \frac{1}{x} - \frac{1}{2 x^2}] + [\ln x + \frac{1}{x} - \frac{1}{2 x^2}] - x \ln x. \] Simplify: \[ E = x \left( \frac{1}{x} - \frac{1}{2 x^2} \right) + \ln x + \frac{1}{x} - \frac{1}{2 x^2} - x \ln x + x \ln x. \] This simplifies to: \[ E = 1 - \frac{1}{2 x} + \ln x + \frac{1}{x} - \frac{1}{2 x^2}. \] Ignoring terms that vanish as \( x \to \infty \), the exponent \( E \) approximates to: \[ E \approx \ln x + 1. \] **Step 4: Compute \( g(x) \)** Now, \( g(x) \) becomes: \[ g(x) = e^{\ln x + 1} = x e. \] **Step 5: Find the Limit** Finally, compute the desired limit: \[ \lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{x e}{x} = e. \] **Answer:** \[ e \] |
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| deepseekv3 | Alright, I have this problem to solve: **Problem:** For every positive real number \( x \), let \( g(x) = \lim_{r \rightarrow 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \). What is the limit of \( \frac{g(x)}{x} \) as \( x \) goes to infinity? Alright, let's break this down. I need to find \( \lim_{x \to \infty} \frac{g(x)}{x} \), where \( g(x) \) is defined as a limit involving \( r \). So, I have nested limits here: first, I need to understand \( g(x) \), and then take the limit of \( \frac{g(x)}{x} \) as \( x \) becomes very large. First, let's tackle \( g(x) = \lim_{r \rightarrow 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \). ### Understanding \( g(x) \) I need to evaluate the limit as \( r \) approaches 0 of \( \left( (x+1)^{r+1} - x^{r+1} \right)^{1/r} \). This looks a bit complicated, so maybe I can simplify the expression inside the limit. Let's denote \( A = (x+1)^{r+1} \) and \( B = x^{r+1} \). So, \( g(x) = \lim_{r \to 0} (A - B)^{1/r} \). But \( A - B \) can be rewritten using the binomial theorem or some approximation, perhaps. Wait, since \( r \) is approaching 0, maybe I can use a first-order approximation for the exponents. Recall that for small \( r \), \( a^{r} \approx 1 + r \ln a \). So, let's apply that to both \( A \) and \( B \): \[ (x+1)^{r+1} = (x+1)^1 \cdot (x+1)^r = (x+1) \cdot (x+1)^r \approx (x+1)(1 + r \ln(x+1)) \] Similarly, \[ x^{r+1} = x \cdot x^r \approx x(1 + r \ln x) \] Now, subtract \( B \) from \( A \): \[ A - B \approx (x+1)(1 + r \ln(x+1)) - x(1 + r \ln x) \] Let's expand this: \[ = (x+1) + r(x+1)\ln(x+1) - x - r x \ln x \] Simplify the constants: \[ = (x + 1 - x) + r(x+1)\ln(x+1) - r x \ln x \] \[ = 1 + r[(x+1)\ln(x+1) - x \ln x] \] So, \( A - B \approx 1 + r \left[ (x+1)\ln(x+1) - x \ln x \right] \). Now, \( g(x) = \lim_{r \to 0} (A - B)^{1/r} \approx \lim_{r \to 0} \left(1 + r \left[ (x+1)\ln(x+1) - x \ln x \right]\right)^{1/r} \). Hmm, this looks like the limit definition of the exponential function. Recall that \( \lim_{r \to 0} (1 + r a)^{1/r} = e^a \). So, applying that here: \[ g(x) \approx e^{(x+1)\ln(x+1) - x \ln x} \] But wait, is this approximation valid? Let's think about it. For small \( r \), the approximation \( a^r \approx 1 + r \ln a \) is reasonable, and the limit \( \lim_{r \to 0} (1 + r a)^{1/r} = e^a \) is a standard result. So, this seems to hold. Now, simplify the exponent: \[ (x+1)\ln(x+1) - x \ln x \] This expression can be simplified further. Let's denote \( y = x \), so: \[ (y+1)\ln(y+1) - y \ln y \] I can expand \( (y+1)\ln(y+1) \) as \( y \ln(y+1) + \ln(y+1) \), so: \[ y \ln(y+1) + \ln(y+1) - y \ln y \] Factor out \( y \): \[ y (\ln(y+1) - \ln y) + \ln(y+1) \] Simplify \( \ln(y+1) - \ln y \): \[ \ln\left(\frac{y+1}{y}\right) = \ln\left(1 + \frac{1}{y}\right) \] So, the expression becomes: \[ y \ln\left(1 + \frac{1}{y}\right) + \ln(y+1) \] As \( y \) (which is \( x \)) goes to infinity, \( \frac{1}{y} \) approaches 0, and \( \ln(y+1) \approx \ln y \). Recall that for large \( y \), \( \ln\left(1 + \frac{1}{y}\right) \approx \frac{1}{y} - \frac{1}{2y^2} + \cdots \). So, the first term becomes: \[ y \left( \frac{1}{y} - \frac{1}{2y^2} + \cdots \right) = 1 - \frac{1}{2y} + \cdots \] And the second term \( \ln(y+1) \approx \ln y \). So, the entire exponent simplifies to: \[ 1 - \frac{1}{2y} + \ln y \] Therefore, \( g(x) \approx e^{1 - \frac{1}{2x} + \ln x} = e^1 \cdot e^{-\frac{1}{2x}} \cdot e^{\ln x} = e \cdot e^{-\frac{1}{2x}} \cdot x \). As \( x \) becomes very large, \( e^{-\frac{1}{2x}} \) approaches 1, so \( g(x) \approx e x \). ### Finding \( \lim_{x \to \infty} \frac{g(x)}{x} \) From the above approximation, \( g(x) \approx e x \), so: \[ \frac{g(x)}{x} \approx \frac{e x}{x} = e \] Therefore, the limit is \( e \). But wait, I should verify if my approximations hold. I used the approximation \( a^r \approx 1 + r \ln a \) for small \( r \), and the limit \( \lim_{r \to 0} (1 + r a)^{1/r} = e^a \). These are standard results, so I think my steps are valid. Additionally, as \( x \) becomes large, \( e^{-\frac{1}{2x}} \) indeed approaches 1, so the approximation \( g(x) \approx e x \) is justified. ### Conclusion After simplifying and taking the limits, I find that: \[ \lim_{x \to \infty} \frac{g(x)}{x} = e \] Therefore, the final answer is: \[ \boxed{e} \] |
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