| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem using Taylor's Approximation Theorem, we need to consider the Taylor series expansions of \( e^\frac{x^4}{2} \) and \( \cos(x^2) \) around \( x=0 \). The Taylor series expansion of \( e^x \) around \( x=0 \) is given by: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \] Substituting \( x^4/2 \) for \( x \), we get the Taylor series expansion of \( e^\frac{x^4}{2} \): \[ e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \frac{(\frac{x^4}{2})^4}{4!} + \cdots \] \[ e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \frac{x^{16}}{384} + \cdots \] The Taylor series expansion of \( \cos(x) \) around \( x=0 \) is given by: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \] Substituting \( x^2 \) for \( x \), we get the Taylor series expansion of \( \cos(x^2) \): \[ \cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \cdots \] \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \] Now, we can substitute the Taylor series expansions of \( e^\frac{x^4}{2} \) and \( \cos(x^2) \) into the given limit: \[ \lim_{x \to 0} \frac{e^\frac{x^4}{2} - \cos(x^2)}{x^4} = \lim_{x \to 0} \frac{\left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots \right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \right)}{x^4} \] Simplifying the expression: \[ \lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots + \frac{x^4}{2} - \frac{x^8}{24} + \frac{x^{12}}{720} - \cdots}{x^4} \] \[ \lim_{x \to 0} \frac{x^4 + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots - \frac{x^8}{24} + \frac{x^{12}}{720} - \cdots}{x^4} \] \[ \lim_{x \to 0} \frac{x^4 + \frac{x^8}{8} - \frac{x^8}{24} + \frac{x^{12}}{48} + \frac{x^{12}}{720} + \cdots}{x^4} \] \[ \lim_{x \to 0} \left(1 + \frac{x^4}{8} - \frac{x^4}{24} + \frac{x^8}{48} + \frac{x^8}{720} + \cdots \right) \] As \( x \to 0 \), all the terms with higher powers of \( x \) go to \( 0 \). Therefore, the limit simplifies to: \[ \lim_{x \to 0} \left(1 + \frac{x^4}{8} - \frac{x^4}{24} + \frac{x^8}{48} + \frac{x^8}{720} + \cdots \right) = 1 + 0 - 0 + 0 + 0 + \cdots = 1 \] Hence, the limit is: \[ \lim_{x \to 0} \frac{e^\frac{x^4}{2} - \cos(x^2)}{x^4} = 1 \] |
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| deepseekr1 | To find the limit \(\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4}\) using Taylor's Approximation Theorem, we start by expanding the functions \(e^{\frac{x^4}{2}}\) and \(\cos(x^2)\) around \(x = 0\). 1. **Taylor series expansion for \(e^{\frac{x^4}{2}}\):** \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{\left(\frac{x^4}{2}\right)^2}{2!} + \cdots = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \cdots \] 2. **Taylor series expansion for \(\cos(x^2)\):** \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \cdots = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \] 3. **Subtracting the two series:** \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \cdots\right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots\right) \] Simplifying the subtraction: \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 - 1\right) + \left(\frac{x^4}{2} + \frac{x^4}{2}\right) + \left(\frac{x^8}{8} - \frac{x^8}{24}\right) + \cdots \] \[ = x^4 + \frac{x^8}{12} + \cdots \] 4. **Dividing by \(x^4\):** \[ \frac{e^\frac{x^4}{2} - \cos(x^2)}{x^4} = \frac{x^4 + \frac{x^8}{12} + \cdots}{x^4} = 1 + \frac{x^4}{12} + \cdots \] 5. **Taking the limit as \(x \to 0\):** \[ \lim_{x \to 0} \left(1 + \frac{x^4}{12} + \cdots\right) = 1 \] Thus, the limit is \(\boxed{1}\). |
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| gpto1 | To find the limit \(\displaystyle \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4}\) using Taylor's Approximation Theorem, we'll expand both \(e^{\frac{x^4}{2}}\) and \(\cos(x^2)\) into their Taylor series around \(x = 0\) and then compute the limit. **Step 1: Expand \(e^{\frac{x^4}{2}}\) using its Taylor series** Recall the Taylor series for \(e^z\) centered at \(z = 0\) is: \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] Substituting \(z = \frac{x^4}{2}\), we get: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{1}{2!} \left( \frac{x^4}{2} \right)^2 + \frac{1}{3!} \left( \frac{x^4}{2} \right)^3 + \cdots \] Simplifying the first few terms: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots \] **Step 2: Expand \(\cos(x^2)\) using its Taylor series** Recall the Taylor series for \(\cos z\) centered at \(z = 0\) is: \[ \cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots \] Substituting \(z = x^2\), we get: \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2} + \frac{(x^2)^4}{24} - \frac{(x^2)^6}{720} + \cdots \] Simplifying the first few terms: \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \] **Step 3: Compute the difference \(e^{\frac{x^4}{2}} - \cos(x^2)\)** Subtract the series of \(\cos(x^2)\) from that of \(e^{\frac{x^4}{2}}\): \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \cdots \right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \right) \] Simplify term by term: 1. The constant terms \(1 - 1 = 0\). 2. The \(x^4\) terms: \[ \frac{x^4}{2} - \left( -\frac{x^4}{2} \right) = \frac{x^4}{2} + \frac{x^4}{2} = x^4 \] 3. The \(x^8\) terms: \[ \frac{x^8}{8} - \frac{x^8}{24} = x^8 \left( \frac{1}{8} - \frac{1}{24} \right) = x^8 \left( \frac{3 - 1}{24} \right) = \frac{x^8}{12} \] So the difference is: \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \frac{x^8}{12} + \cdots \] **Step 4: Divide by \(x^4\) and compute the limit** Divide the difference by \(x^4\): \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 + \frac{x^4}{12} + \cdots \] As \(x \to 0\), the higher-order terms involving \(x^4\) and beyond vanish: \[ \lim_{x \to 0} \left( 1 + \frac{x^4}{12} + \cdots \right) = 1 \] **Conclusion:** \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 \] **Answer:** 1 |
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| GPT-4o | To solve the limit \(\lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4}\) using Taylor's Approximation Theorem, let's start by expanding \(e^{\frac{x^4}{2}}\) and \(\cos(x^2)\) into their Taylor series around \(x = 0\). ### Step 1: Taylor Series Expansion 1. **Taylor Series for \(e^{y}\):** \[ e^{y} = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \] Set \(y = \frac{x^4}{2}\): \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{\left(\frac{x^4}{2}\right)^2}{2!} + \frac{\left(\frac{x^4}{2}\right)^3}{3!} + \cdots \] \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + O(x^{12}) \] 2. **Taylor Series for \(\cos(z)\):** \[ \cos(z) = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots \] Set \(z = x^2\): \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \frac{(x^2)^6}{6!} + \cdots \] \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} + O(x^{12}) \] ### Step 2: Subtract the Series Now, subtract the Taylor series of \(\cos(x^2)\) from \(e^{\frac{x^4}{2}}\): \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + O(x^{12})\right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} + O(x^{12})\right) \] Simplify the subtraction: \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8}\right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24}\right) + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = 1 + \frac{x^4}{2} + \frac{x^8}{8} - 1 + \frac{x^4}{2} - \frac{x^8}{24} + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = \frac{x^4}{2} + \frac{x^4}{2} + \frac{x^8}{8} - \frac{x^8}{24} + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \left(\frac{x^8}{8} - \frac{x^8}{24}\right) + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \left(\frac{3x^8}{24} - \frac{x^8}{24}\right) + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \frac{2x^8}{24} + O(x^{12}) \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \frac{x^8}{12} + O(x^{12}) \] ### Step 3: Divide by \(x^4\) Now, divide the result by \(x^4\): \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = \frac{x^4 + \frac{x^8}{12} + O(x^{12})}{x^ |
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| llama3.1-70B-Instruct | ## Step 1: Recall Taylor's Approximation Theorem Taylor's Approximation Theorem states that for a function f(x) that is infinitely differentiable at x = a, the value of the function at x = a + h can be approximated by the Taylor series: f(a + h) = f(a) + hf'(a) + (h^2/2!)f''(a) + (h^3/3!)f'''(a) + ... + (h^n/n!)f^(n)(a) + R_n(h), where R_n(h) is the remainder term. ## Step 2: Apply Taylor's Approximation Theorem to e^(x^4/2) and cos(x^2) Using Taylor's Approximation Theorem, we can expand e^(x^4/2) and cos(x^2) around x = 0. The Taylor series expansion of e^(x^4/2) is 1 + (x^4/2) + (x^8/8) + ..., and the Taylor series expansion of cos(x^2) is 1 - (x^4/2) + (x^8/24) - .... ## Step 3: Subtract the Taylor series expansions of cos(x^2) from e^(x^4/2) Subtracting the Taylor series expansions of cos(x^2) from e^(x^4/2), we get: e^(x^4/2) - cos(x^2) = (x^4/2) + (x^8/8) - (x^4/2) + (x^8/24) - ... = (x^8/8) + (x^8/24) + ... = (5x^8/24) + ... ## Step 4: Divide the result by x^4 Dividing the result by x^4, we get: (e^(x^4/2) - cos(x^2))/x^4 = (5x^8/24x^4) + ... = (5x^4/24) + ... ## Step 5: Take the limit as x approaches 0 Taking the limit as x approaches 0, we get: lim (x→0) (e^(x^4/2) - cos(x^2))/x^4 = lim (x→0) (5x^4/24) + ... = 0 + ... = 0. The final answer is: $\boxed{0}$ |
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| llama3.1-405B-Instruct | We can use Taylor's Approximation Theorem to find the limit. First, we can write the Taylor series for $e^\frac{x^4}{2}$ and $\cos(x^2)$: $$e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots$$ $$\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots$$ Now, we can substitute these series into the expression: $$\frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \frac{1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots - (1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots)}{x^4}$$ Simplifying, we get: $$\frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \frac{x^4 + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots + \frac{x^4}{2} - \frac{x^8}{24} + \frac{x^{12}}{720} + \cdots}{x^4}$$ Combining like terms, we get: $$\frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \frac{3x^4}{2x^4} + \frac{x^8}{8x^4} + \frac{x^{12}}{48x^4} + \cdots$$ Now, we can take the limit as $x \to 0$: $$\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \lim_{x \to 0} \left(\frac{3}{2} + \frac{x^4}{8} + \frac{x^8}{48} + \cdots\right)$$ As $x \to 0$, all the terms except the first one go to zero, so we are left with: $$\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \frac{3}{2}$$ Therefore, the limit is $\boxed{\frac{3}{2}}$. |
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| DeepSeek-Coder-V2 | To solve the problem using Taylor's Approximation Theorem, we need to expand both \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) around \( x = 0 \) up to the necessary order. First, let's expand \( e^{\frac{x^4}{2}} \) around \( x = 0 \): \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{1}{2!} \left( \frac{x^4}{2} \right)^2 + \frac{1}{3!} \left( \frac{x^4}{2} \right)^3 + \cdots \] Since we are interested in the limit as \( x \to 0 \), we can truncate the series at the quadratic term: \[ e^{\frac{x^4}{2}} \approx 1 + \frac{x^4}{2} \] Next, let's expand \( \cos(x^2) \) around \( x = 0 \): \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \cdots \] Again, truncating the series at the quadratic term: \[ \cos(x^2) \approx 1 - \frac{x^4}{2} \] Now, we substitute these approximations into the original limit expression: \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} \] \[ = \lim_{x \to 0} \frac{\left(1 + \frac{x^4}{2}\right) - \left(1 - \frac{x^4}{2}\right)}{x^4} \] \[ = \lim_{x \to 0} \frac{1 + \frac{x^4}{2} - 1 + \frac{x^4}{2}}{x^4} \] \[ = \lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^4}{2}}{x^4} \] \[ = \lim_{x \to 0} \frac{x^4}{x^4} \] \[ = \lim_{x \to 0} 1 \] \[ = 1 \] Thus, the limit is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | To solve the problem using Taylor's Approximation Theorem, we need to consider the Taylor series expansions of $e^\frac{x^4}{2}$ and $\cos(x^2)$ around $x=0$. The Taylor series expansion of $e^x$ around $x=0$ is given by: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ Substituting $x^4/2$ for $x$, we get the Taylor series expansion of $e^\frac{x^4}{2}$: $$e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \frac{(\frac{x^4}{2})^4}{4!} + \cdots$$ $$e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \frac{x^{16}}{384} + \cdots$$ The Taylor series expansion of $\cos(x)$ around $x=0$ is given by: $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ Substituting $x^2$ for $x$, we get the Taylor series expansion of $\cos(x^2)$: $$\cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \cdots$$ $$\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots$$ Now, we can substitute the Taylor series expansions of $e^\frac{x^4}{2}$ and $\cos(x^2)$ into the given limit: $$\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = \lim_{x \to 0} \frac{\left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots\right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots\right)}{x^4}$$ Simplifying the expression: $$\lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots + \frac{x^4}{2} - \frac{x^8}{24} + \frac{x^{12}}{720} - \cdots}{x^4}$$ $$\lim_{x \to 0} \frac{x^4 + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots - \frac{x^8}{24} + \frac{x^{12}}{720} - \cdots}{x^4}$$ $$\lim_{x \to 0} \frac{x^4 + \frac{x^8}{8} - \frac{x^8}{24} + \frac{x^{12}}{48} + \frac{x^{12}}{720} + \cdots}{x^4}$$ $$\lim_{x \to 0} \left(1 + \frac{x^4}{8} - \frac{x^4}{24} + \frac{x^8}{48} + \frac{x^8}{720} + \cdots\right)$$ As $x \to 0$, all the terms with higher powers of $x$ go to $0$. Therefore, the limit simplifies to: $$\lim_{x \to 0} \left(1 + \frac{x^4}{8} - \frac{x^4}{24} + \frac{x^8}{48} + \frac{x^8}{720} + \cdots\right) = 1 + 0 - 0 + 0 + 0 + \cdots = 1$$ Hence, the limit is: $$\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = 1$$ |
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| baichuan4 | To find the limit $\lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4}$ using Taylor's Approximation Theorem, we can expand the exponential and cosine functions into their Taylor series around $x=0$. The Taylor series for $e^x$ at $x=0$ is: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + O(x^5)$$ Substituting $\frac{x^4}{2}$ for $x$, we get: $$e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \frac{(\frac{x^4}{2})^4}{4!} + O(x^{10})$$ The Taylor series for $\cos(x)$ at $x=0$ is: $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + O(x^8)$$ Substituting $x^2$ for $x$, we get: $$\cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + O(x^{16})$$ Now, let's compute the limit by substituting these series into the original expression: $$\lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = \lim_{x \to 0} \frac{\left(1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \frac{(\frac{x^4}{2})^4}{4!} + O(x^{10})\right) - \left(1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + O(x^{16})\right)}{x^4}$$ Simplifying and canceling out the leading terms: $$= \lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} - \frac{x^8}{4!} + \frac{(\frac{x^4}{2})^3}{3!} - O(x^{10})}{x^4}$$ Since we are interested in the limit as $x$ approaches $0$, we can neglect the higher-order terms (as they will go to $0$ faster than the lower-order terms) and focus on the lowest-order non-zero terms: $$= \lim_{x \to 0} \frac{\frac{x^4}{2}}{x^4}$$ This simplifies to: $$= \frac{1}{2}$$ Therefore, the limit $\lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4}$ is equal to $\frac{1}{2}$. |
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| doubao | Let's find the Taylor series expansions of $e^{\frac{x^4}{2}}$ and $\cos(x^2)$ near $x=0$. The Taylor series expansion of $e^t$ is $1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$. So for $e^{\frac{x^4}{2}}$, we have: $e^{\frac{x^4}{2}}=1+\frac{x^4}{2}+\frac{(\frac{x^4}{2})^2}{2!}+\frac{(\frac{x^4}{2})^3}{3!}+\cdots$. The Taylor series expansion of $\cos t$ is $1-\frac{t^2}{2!}+\frac{t^4}{4!}-\cdots$. So for $\cos(x^2)$, we have: $\cos(x^2)=1-\frac{(x^2)^2}{2!}+\frac{(x^2)^4}{4!}-\cdots=1-\frac{x^4}{2}+\frac{x^8}{24}-\cdots$. Now, substitute these expansions into the given expression: \[ \begin{align*} &\lim_{x \to 0} \frac{e^{\frac{x^4}{2}}-\cos(x^2)}{x^4}\\ =&\lim_{x \to 0} \frac{(1+\frac{x^4}{2}+\frac{(\frac{x^4}{2})^2}{2!}+\frac{(\frac{x^4}{2})^3}{3!}+\cdots)-(1-\frac{x^4}{2}+\frac{x^8}{24}-\cdots)}{x^4}\\ =&\lim_{x \to 0} \frac{\frac{x^4}{2}+\frac{x^8}{8}+\cdots+\frac{x^4}{2}-\frac{x^8}{24}+\cdots}{x^4}\\ =&\lim_{x \to 0} \frac{x^4+\frac{x^8}{4}+\cdots}{x^4}\\ =&1+\lim_{x \to 0} \frac{x^4}{4}+\cdots\\ =&1. \end{align*} \] |
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| minimax | To solve the given limit using Taylor's Approximation Theorem, we need to expand both \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) into their Taylor series expansions around \( x = 0 \) and then evaluate the limit. First, let's recall the Taylor series expansions for the exponential function and the cosine function around \( x = 0 \): For \( e^z \), where \( z \) is a general variable, the Taylor series expansion around \( z = 0 \) is: \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots \] For \( \cos(z) \), the Taylor series expansion around \( z = 0 \) is: \[ \cos(z) = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \] Now, let's expand \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) using these series: For \( e^{\frac{x^4}{2}} \), we substitute \( z = \frac{x^4}{2} \) into the exponential series: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \frac{(\frac{x^4}{2})^4}{4!} + \cdots \] \[ = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \frac{x^{16}}{384} + \cdots \] For \( \cos(x^2) \), we substitute \( z = x^2 \) into the cosine series: \[ \cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots \] \[ = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \] Now, let's subtract the series for \( \cos(x^2) \) from the series for \( e^{\frac{x^4}{2}} \) and only keep terms up to \( x^4 \) since higher-order terms will not contribute to the limit as \( x \to 0 \): \[ e^{\frac{x^4}{2}} - \cos(x^2) \approx \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \cdots \right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \right) \] \[ = \frac{x^4}{2} + \frac{x^4}{2} + \frac{x^8}{8} - \frac{x^8}{24} + \cdots \] \[ = x^4 + \frac{x^8}{8} - \frac{x^8}{24} + \cdots \] \[ = x^4 + \frac{3x^8}{24} + \cdots \] Now, we can see that the \( x^4 \) term is the leading term in the expansion, and all other terms are higher order in \( x \). Therefore, when we divide by \( x^4 \), the higher-order terms will vanish in the limit as \( x \to 0 \): \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = \lim_{x \to 0} \frac{x^4 + \frac{3x^8}{24} + \cdots}{x^4} \] \[ = \lim_{x \to 0} \left(1 + \frac{3x^4}{24} + \cdots \right) \] \[ = 1 \] So, the limit is \( \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 \). |
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| kimi | To solve this limit using Taylor's Approximation Theorem, we need to find the Taylor series expansions for \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) around \( x = 0 \). The Taylor series expansion of a function \( f(x) \) around \( x = a \) is given by: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \] Let's find the Taylor series for \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) around \( x = 0 \). For \( e^{\frac{x^4}{2}} \): The Taylor series for \( e^x \) is \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \). So, for \( e^{\frac{x^4}{2}} \), we substitute \( x \) with \( \frac{x^4}{2} \): \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \cdots \] For \( \cos(x^2) \): The Taylor series for \( \cos(x) \) is \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \). So, for \( \cos(x^2) \), we substitute \( x \) with \( x^2 \): \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \frac{(x^2)^6}{6!} + \cdots \] Now, let's find the limit: \[ \lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} \] Using the Taylor series expansions: \[ \lim_{x \to 0} \frac{1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \cdots - (1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots)}{x^4} \] Simplifying the expression: \[ \lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^8}{8} + \cdots + \frac{x^4}{2!} - \frac{x^8}{4!} + \cdots}{x^4} \] \[ \lim_{x \to 0} \left( \frac{1}{2} + \frac{x^4}{8} + \cdots + \frac{1}{2} - \frac{x^4}{4!} + \cdots \right) \] As \( x \to 0 \), all terms with \( x \) in them go to 0. So, we are left with: \[ \lim_{x \to 0} \left( \frac{1}{2} + \frac{1}{2} \right) = 1 \] Therefore, the limit is: \[ \boxed{1} \] |
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| wenxinyiyan | To solve this problem, we will use Taylor's Approximation Theorem to expand both $e^\frac{x^4}{2}$ and $\cos(x^2)$ around $x=0$ and then evaluate the limit. First, let's expand $e^\frac{x^4}{2}$ around $x=0$. The Taylor series expansion of $e^u$ around $u=0$ is given by $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$. Substituting $u = \frac{x^4}{2}$, we get: $$e^\frac{x^4}{2} = 1 + \frac{x^4}{2} + \frac{(\frac{x^4}{2})^2}{2!} + \frac{(\frac{x^4}{2})^3}{3!} + \cdots$$ $$= 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots$$ Next, let's expand $\cos(x^2)$ around $x=0$. The Taylor series expansion of $\cos(u)$ around $u=0$ is given by $\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots$. Substituting $u = x^2$, we get: $$\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots$$ Now, let's evaluate the limit: $$\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4}$$ $$= \lim_{x \to 0} \frac{(1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots) - (1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots)}{x^4}$$ $$= \lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^4}{2} + \frac{x^8}{8} - \frac{x^8}{24} + \frac{x^{12}}{48} + \frac{x^{12}}{720} + \cdots}{x^4}$$ $$= \lim_{x \to 0} \left( 1 + \frac{x^4}{4} - \frac{x^8}{12} + \frac{x^{12}}{24} + \cdots \right)$$ $$= 1$$ Therefore, $\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4} = 1$. |
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| zhipu | To solve the limit \(\lim_{x \to 0} \frac{e^\frac{x^4}{2}-\cos(x^2)}{x^4}\) using Taylor's Approximation Theorem, we will expand the functions \(e^{\frac{x^4}{2}}\) and \(\cos(x^2)\) into their Taylor series around \(x = 0\). ### Step 1: Taylor Series Expansion of \(e^{\frac{x^4}{2}}\) The Taylor series expansion of \(e^u\) around \(u = 0\) is: \[ e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots \] For \(u = \frac{x^4}{2}\), we get: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{\left(\frac{x^4}{2}\right)^2}{2!} + \frac{\left(\frac{x^4}{2}\right)^3}{3!} + \cdots \] \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots \] ### Step 2: Taylor Series Expansion of \(\cos(x^2)\) The Taylor series expansion of \(\cos(u)\) around \(u = 0\) is: \[ \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots \] For \(u = x^2\), we get: \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \frac{(x^2)^6}{6!} + \cdots \] \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \] ### Step 3: Subtract the Series Now, subtract the Taylor series of \(\cos(x^2)\) from \(e^{\frac{x^4}{2}}\): \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots \right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \right) \] Simplify the expression: \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 - 1\right) + \left(\frac{x^4}{2} + \frac{x^4}{2}\right) + \left(\frac{x^8}{8} - \frac{x^8}{24}\right) + \left(\frac{x^{12}}{48} + \frac{x^{12}}{720}\right) + \cdots \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \left(\frac{x^8}{8} - \frac{x^8}{24}\right) + \left(\frac{x^{12}}{48} + \frac{x^{12}}{720}\right) + \cdots \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \left(\frac{3x^8}{24}\right) + \left(\frac{19x^{12}}{1440}\right) + \cdots \] \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \frac{x^8}{8} + \cdots \] ### Step 4: Form the Limit Expression Now, form the limit expression: \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = \frac{x^4 + \frac{x^8}{8} + \cdots}{x^4} \] \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 + \frac{x^4}{8} + \cdots \] As \(x \to 0\), the higher-order terms \(\frac{x^4}{8} + \cdots\) go to 0. Thus, the limit is: \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 \] ### Conclusion \[ \boxed{1} \] |
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| gpto1 | To find the limit \(\displaystyle \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4}\) using Taylor's Approximation Theorem, we'll expand both \(e^{\frac{x^4}{2}}\) and \(\cos(x^2)\) into their Taylor series around \(x = 0\) and then compute the limit. **Step 1: Expand \(e^{\frac{x^4}{2}}\) using its Taylor series** Recall the Taylor series for \(e^z\) centered at \(z = 0\) is: \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] Substituting \(z = \frac{x^4}{2}\), we get: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{1}{2!} \left( \frac{x^4}{2} \right)^2 + \frac{1}{3!} \left( \frac{x^4}{2} \right)^3 + \cdots \] Simplifying the first few terms: \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{x^8}{8} + \frac{x^{12}}{48} + \cdots \] **Step 2: Expand \(\cos(x^2)\) using its Taylor series** Recall the Taylor series for \(\cos z\) centered at \(z = 0\) is: \[ \cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots \] Substituting \(z = x^2\), we get: \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2} + \frac{(x^2)^4}{24} - \frac{(x^2)^6}{720} + \cdots \] Simplifying the first few terms: \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots \] **Step 3: Compute the difference \(e^{\frac{x^4}{2}} - \cos(x^2)\)** Subtract the series of \(\cos(x^2)\) from that of \(e^{\frac{x^4}{2}}\): \[ e^{\frac{x^4}{2}} - \cos(x^2) = \left(1 + \frac{x^4}{2} + \frac{x^8}{8} + \cdots \right) - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \right) \] Simplify term by term: 1. The constant terms \(1 - 1 = 0\). 2. The \(x^4\) terms: \[ \frac{x^4}{2} - \left( -\frac{x^4}{2} \right) = \frac{x^4}{2} + \frac{x^4}{2} = x^4 \] 3. The \(x^8\) terms: \[ \frac{x^8}{8} - \frac{x^8}{24} = x^8 \left( \frac{1}{8} - \frac{1}{24} \right) = x^8 \left( \frac{3 - 1}{24} \right) = \frac{x^8}{12} \] So the difference is: \[ e^{\frac{x^4}{2}} - \cos(x^2) = x^4 + \frac{x^8}{12} + \cdots \] **Step 4: Divide by \(x^4\) and compute the limit** Divide the difference by \(x^4\): \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 + \frac{x^4}{12} + \cdots \] As \(x \to 0\), the higher-order terms involving \(x^4\) and beyond vanish: \[ \lim_{x \to 0} \left( 1 + \frac{x^4}{12} + \cdots \right) = 1 \] **Conclusion:** \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 \] **Answer:** 1 |
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| deepseekv3 | To evaluate the limit: \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} \] we can use **Taylor's Approximation Theorem** to expand the functions \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) around \( x = 0 \). --- ### Step 1: Taylor Expansion of \( e^{\frac{x^4}{2}} \) The Taylor series for \( e^t \) around \( t = 0 \) is: \[ e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots \] Substitute \( t = \frac{x^4}{2} \): \[ e^{\frac{x^4}{2}} = 1 + \frac{x^4}{2} + \frac{\left(\frac{x^4}{2}\right)^2}{2!} + \cdots \] For small \( x \), we can truncate the series after the \( x^4 \) term: \[ e^{\frac{x^4}{2}} \approx 1 + \frac{x^4}{2} \] --- ### Step 2: Taylor Expansion of \( \cos(x^2) \) The Taylor series for \( \cos(u) \) around \( u = 0 \) is: \[ \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \] Substitute \( u = x^2 \): \[ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots \] For small \( x \), we can truncate the series after the \( x^4 \) term: \[ \cos(x^2) \approx 1 - \frac{x^4}{2} \] --- ### Step 3: Substitute the Expansions into the Limit Substitute the Taylor approximations of \( e^{\frac{x^4}{2}} \) and \( \cos(x^2) \) into the limit: \[ \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} \approx \frac{\left(1 + \frac{x^4}{2}\right) - \left(1 - \frac{x^4}{2}\right)}{x^4} \] Simplify the numerator: \[ \frac{\left(1 + \frac{x^4}{2}\right) - \left(1 - \frac{x^4}{2}\right)}{x^4} = \frac{1 + \frac{x^4}{2} - 1 + \frac{x^4}{2}}{x^4} \] \[ = \frac{\frac{x^4}{2} + \frac{x^4}{2}}{x^4} \] \[ = \frac{x^4}{x^4} \] \[ = 1 \] --- ### Final Answer: \[ \lim_{x \to 0} \frac{e^{\frac{x^4}{2}} - \cos(x^2)}{x^4} = 1 \] |
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