| 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 \(\lim_{x \to 0} \frac{\cos(mx - 1)}{x^2}\) and see if it equals \(-\frac{m^2}{2}\) when \(m = 2\). First, let's rewrite the limit in a more manageable form: \[ \lim_{x \to 0} \frac{\cos(mx - 1)}{x^2} \] We know that \(\cos(mx - 1)\) can be expanded using the Taylor series around \(x = 0\). However, a more straightforward approach is to use L'Hopital's rule since the limit is of the form \(\frac{0}{0}\). Applying L'Hopital's rule, we differentiate the numerator and the denominator: \[ \lim_{x \to 0} \frac{\cos(mx - 1)}{x^2} = \lim_{x \to 0} \frac{\frac{d}{dx}[\cos(mx - 1)]}{\frac{d}{dx}[x^2]} \] The derivative of \(\cos(mx - 1)\) with respect to \(x\) is: \[ \frac{d}{dx}[\cos(mx - 1)] = -m \sin(mx - 1) \] And the derivative of \(x^2\) is: \[ \frac{d}{dx}[x^2] = 2x \] So, we have: \[ \lim_{x \to 0} \frac{-m \sin(mx - 1)}{2x} \] Applying L'Hopital's rule again since this is still of the form \(\frac{0}{0}\): \[ \lim_{x \to 0} \frac{-m \sin(mx - 1)}{2x} = \lim_{x \to 0} \frac{\frac{d}{dx}[-m \sin(mx - 1)]}{\frac{d}{dx}[2x]} \] The derivative of \(-m \sin(mx - 1)\) with respect to \(x\) is: \[ \frac{d}{dx}[-m \sin(mx - 1)] = -m^2 \cos(mx - 1) \] And the derivative of \(2x\) is: \[ \frac{d}{dx}[2x] = 2 \] So, we have: \[ \lim_{x \to 0} \frac{-m^2 \cos(mx - 1)}{2} \] As \(x \to 0\), \(\cos(mx - 1) \to \cos(-1)\), which is a constant. Therefore: \[ \lim_{x \to 0} \frac{-m^2 \cos(-1)}{2} = -\frac{m^2 \cos(-1)}{2} \] For \(m = 2\), we get: \[ -\frac{2^2 \cos(-1)}{2} = -2 \cos(-1) \] Since \(\cos(-1)\) is a constant and not equal to 1, the limit does not equal \(-\frac{m^2}{2}\) for \(m = 2\). Therefore, the answer is no, the limit \(\lim_{x \to 0} \frac{\cos(mx - 1)}{x^2}\) does not equal \(-\frac{m^2}{2}\) for \(m = 2\). |
True |
| deepseekr1 | To determine if \(\lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{2^2}{2}\), we start by analyzing the given expression. First, we substitute \(m = 2\) into the expression, resulting in \(\frac{\cos(2x) - 1}{x^2}\). Using the Taylor series expansion for \(\cos(2x)\) around \(x = 0\): \[ \cos(2x) = 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} - \cdots \] Subtracting 1 and dividing by \(x^2\): \[ \frac{\cos(2x) - 1}{x^2} = \frac{-2x^2 + \frac{16x^4}{24} - \cdots}{x^2} = -2 + \frac{16x^2}{24} - \cdots \] As \(x \to 0\), the higher-order terms vanish, leaving the limit as \(-2\). We can also use L'Hospital's Rule. The initial form is \(\frac{0}{0}\), so we apply the rule twice: 1. First derivative: \(\frac{-2\sin(2x)}{2x} = \frac{-\sin(2x)}{x}\) 2. Second derivative: \(\frac{-2\cos(2x)}{1}\) Evaluating the second derivative at \(x = 0\) gives \(-2\cos(0) = -2\). Both methods confirm the limit is \(-2\), which is \(-\frac{2^2}{2} = -2\). \[ \boxed{-2} \] |
True |
| gpto1 | **Solution:** We are asked to determine whether the following limit holds true for \( m = 2 \): \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \] Specifically, we need to compute: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \] **Step 1: Understanding the Limit** The function involves the cosine function and a quadratic term in the denominator. As \( x \to 0 \), both the numerator and the denominator approach zero: - Numerator: \( \cos(2x) - 1 \to \cos(0) - 1 = 1 - 1 = 0 \) - Denominator: \( x^2 \to 0^2 = 0 \) Since we have an indeterminate form \( \frac{0}{0} \), we can use techniques like the Taylor series expansion or L'Hôpital's Rule to evaluate the limit. --- **Method 1: Using the Taylor Series Expansion** The Taylor series expansion of \( \cos(u) \) around \( u = 0 \) is: \[ \cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \frac{u^6}{720} + \cdots \] For very small \( x \), higher-order terms become negligible. Therefore, we can approximate \( \cos(2x) \) as: \[ \cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - \frac{4x^2}{2} = 1 - 2x^2 \] **Substituting into the limit:** \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \approx \lim_{x \to 0} \frac{(1 - 2x^2) - 1}{x^2} = \lim_{x \to 0} \frac{-2x^2}{x^2} = \lim_{x \to 0} -2 = -2 \] Thus, we find: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] Since \( m = 2 \), we compute \( -\frac{m^2}{2} \): \[ -\frac{m^2}{2} = -\frac{(2)^2}{2} = -\frac{4}{2} = -2 \] **Conclusion from Method 1:** The limit equals \( -2 \), which matches \( -\frac{m^2}{2} \) when \( m = 2 \). --- **Method 2: Using L'Hôpital's Rule** Since the limit is in the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to \( x \). **First Application of L'Hôpital's Rule:** Let: - Numerator \( N(x) = \cos(2x) - 1 \) - Denominator \( D(x) = x^2 \) Compute the derivatives: - \( N'(x) = -2\sin(2x) \) - \( D'(x) = 2x \) Apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = \lim_{x \to 0} \frac{N'(x)}{D'(x)} = \lim_{x \to 0} \frac{-2\sin(2x)}{2x} = \lim_{x \to 0} \frac{-\sin(2x)}{x} \] **Second Application of L'Hôpital's Rule:** Again, the limit is in the indeterminate form \( \frac{0}{0} \): - Numerator \( N_1(x) = -\sin(2x) \) - Denominator \( D_1(x) = x \) Compute the derivatives: - \( N_1'(x) = -2\cos(2x) \) - \( D_1'(x) = 1 \) Apply L'Hôpital's Rule again: \[ \lim_{x \to 0} \frac{-\sin(2x)}{x} = \lim_{x \to 0} \frac{N_1'(x)}{D_1'(x)} = \lim_{x \to 0} \frac{-2\cos(2x)}{1} = -2\cos(0) = -2(1) = -2 \] **Conclusion from Method 2:** Again, we find: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 = -\frac{(2)^2}{2} \] --- **Final Answer:** Yes, the limit holds true for \( m = 2 \): \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{(2)^2}{2} = -2 \] **Therefore,** \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \quad \text{when} \quad m = 2 \] |
True |
| GPT-4o | To solve the problem \(\lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2}\) for \(m = 2\), we need to evaluate the limit: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \] We'll use the Taylor series expansion of \(\cos(2x)\) around \(x = 0\): \[ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \cdots \] For small \(x\), the higher-order terms become negligible, so we can approximate \(\cos(2x)\) by truncating the series after the second term: \[ \cos(2x) \approx 1 - \frac{(2x)^2}{2} \] Simplifying, we get: \[ \cos(2x) \approx 1 - 2x^2 \] Now, substitute this approximation into the limit: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \approx \lim_{x \to 0} \frac{(1 - 2x^2) - 1}{x^2} = \lim_{x \to 0} \frac{-2x^2}{x^2} = \lim_{x \to 0} -2 = -2 \] Therefore, we have shown that: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] This matches the expression \(-\frac{m^2}{2}\) given in the problem for \(m = 2\): \[ -\frac{2^2}{2} = -\frac{4}{2} = -2 \] Thus, the solution is verified, and the limit is indeed: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] |
True |
| llama3.1-70B-Instruct | To solve this problem, we can use L'Hopital's rule, which states that if a limit is in the form 0/0 or ∞/∞, we can differentiate the numerator and denominator separately and then take the limit. In this case, we have: lim (x→0) (cos(mx - 1)/(x^2)) We can rewrite this as: lim (x→0) (cos(mx - 1))/(x^2) Since the limit is in the form 0/0, we can apply L'Hopital's rule: lim (x→0) (d(cos(mx - 1))/dx) / (d(x^2)/dx) = lim (x→0) (-m sin(mx - 1)) / (2x) Now, we can substitute m = 2: lim (x→0) (-2 sin(2x - 1)) / (2x) = lim (x→0) (-sin(2x - 1)) / x Using the fact that sin(u) ≈ u for small values of u, we can approximate: lim (x→0) (-sin(2x - 1)) / x ≈ lim (x→0) (-2x) / x = -2 However, we need to be careful here. The original limit was in the form 0/0, and we applied L'Hopital's rule to get a new limit. But the new limit is not necessarily equal to the original limit. To confirm that the limit is indeed -2, we can use the Taylor series expansion of cos(mx - 1): cos(mx - 1) = 1 - (mx - 1)^2/2! + ... Substituting m = 2, we get: cos(2x - 1) = 1 - (2x - 1)^2/2! + ... = 1 - (4x^2 - 4x + 1)/2 + ... = 1 - 2x^2 + 2x - 1/2 + ... Now, we can substitute this back into the original limit: lim (x→0) (cos(2x - 1))/(x^2) = lim (x→0) (1 - 2x^2 + 2x - 1/2 + ...)/(x^2) = lim (x→0) (-2 + 2/x - 1/2 + ...) = -2 Therefore, the limit is indeed -2, which is equal to -(m^2)/2 for m = 2. The final answer is -2. |
True |
| llama3.1-405B-Instruct | To determine if the limit \lim_{x \to 0} (cos(mx - 1)/(x^2) = -(m^2)/2 for m = 2, we can use L'Hopital's rule. First, let's substitute m = 2 into the expression: \lim_{x \to 0} (cos(2x - 1)/(x^2) We can simplify the expression by using the trigonometric identity cos(2x) = 1 - 2sin^2(x): \lim_{x \to 0} (1 - 2sin^2(x) - 1)/(x^2) Simplifying further, we get: \lim_{x \to 0} (-2sin^2(x))/(x^2) Now, we can apply L'Hopital's rule, which states that if a limit is in the form 0/0, we can differentiate the numerator and denominator separately and then take the limit again. Differentiating the numerator and denominator, we get: \lim_{x \to 0} (-4sin(x)cos(x))/(2x) Simplifying, we get: \lim_{x \to 0} (-2sin(x)cos(x))/x Now, we can apply L'Hopital's rule again: \lim_{x \to 0} (-2cos^2(x) + 2sin^2(x))/1 Simplifying, we get: \lim_{x \to 0} (-2cos^2(x) + 2sin^2(x)) Evaluating the limit, we get: -2(1) + 2(0) = -2 Therefore, the limit \lim_{x \to 0} (cos(mx - 1)/(x^2) = -(m^2)/2 for m = 2 is true. |
True |
| DeepSeek-Coder-V2 | To solve the given problem, we need to evaluate the limit of the function \(\frac{\cos(mx - 1)}{x^2}\) as \(x\) approaches 0, and determine if it equals \(-\frac{m^2}{2}\) for \(m = 2\). Let's proceed step by step: 1. **Substitute \(m = 2\) into the function:** \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \] 2. **Use the Taylor series expansion for \(\cos(2x - 1)\) around \(x = 0\):** The Taylor series for \(\cos(u)\) around \(u = -1\) is: \[ \cos(u) = \cos(-1) - \sin(-1)(u + 1) + \frac{\cos(-1)}{2}(u + 1)^2 + \cdots \] Substituting \(u = 2x - 1\): \[ \cos(2x - 1) = \cos(-1) - \sin(-1)(2x) + \frac{\cos(-1)}{2}(2x)^2 + \cdots \] Simplifying: \[ \cos(2x - 1) = \cos(-1) - 2x\sin(-1) + 2x^2\cos(-1) + \cdots \] 3. **Substitute the Taylor series into the limit:** \[ \lim_{x \to 0} \frac{\cos(-1) - 2x\sin(-1) + 2x^2\cos(-1) + \cdots}{x^2} \] 4. **Separate the terms in the numerator:** \[ \lim_{x \to 0} \left( \frac{\cos(-1)}{x^2} - \frac{2x\sin(-1)}{x^2} + \frac{2x^2\cos(-1)}{x^2} + \cdots \right) \] Simplifying each term: \[ \lim_{x \to 0} \left( \frac{\cos(-1)}{x^2} - \frac{2\sin(-1)}{x} + 2\cos(-1) + \cdots \right) \] 5. **Evaluate the limit term by term:** - The term \(\frac{\cos(-1)}{x^2}\) diverges as \(x \to 0\). - The term \(\frac{2\sin(-1)}{x}\) also diverges as \(x \to 0\). - The term \(2\cos(-1)\) is a constant and does not affect the limit. Since the terms involving \(x\) in the denominator diverge, the limit does not exist in the conventional sense. Therefore, the given statement is not valid. **Conclusion:** The limit \(\lim_{x \to 0} \frac{\cos(2x - 1)}{x^2}\) does not equal \(-\frac{2^2}{2} = -2\). Thus, the answer is: \[ \boxed{\text{False}} \] |
True |
| Qwen2-72B-Instruct | To solve this problem, we need to evaluate the limit \(\lim_{x \to 0} \frac{\cos(mx) - 1}{x^2}\) and see if it equals \(-\frac{m^2}{2}\) when \(m = 2\). First, let's apply the limit directly: \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} \] We know that \(\cos(mx) = 1 - \frac{(mx)^2}{2!} + \frac{(mx)^4}{4!} - \ldots\) from the Taylor series expansion of cosine. So, we can rewrite the limit as: \[ \lim_{x \to 0} \frac{1 - \frac{(mx)^2}{2!} + \ldots - 1}{x^2} \] \[ = \lim_{x \to 0} \frac{-\frac{(mx)^2}{2!}}{x^2} \] \[ = \lim_{x \to 0} -\frac{m^2x^2}{2x^2} \] \[ = \lim_{x \to 0} -\frac{m^2}{2} \] \[ = -\frac{m^2}{2} \] So, for \(m = 2\), we have: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{2^2}{2} = -2 \] Therefore, the solution is yes, the limit \(\lim_{x \to 0} \frac{\cos(mx) - 1}{x^2}\) does equal \(-\frac{m^2}{2}\) for \(m = 2\). |
True |
| baichuan4 | To solve the problem, we need to evaluate the limit: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \] We can use L'Hôpital's Rule because the form of the limit is \(\frac{0}{0}\) as \(x\) approaches 0. L'Hôpital's Rule states that if you have an indeterminate form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you can take the derivative of the numerator and the denominator separately and the limit will be the same. So, let's differentiate the numerator and the denominator with respect to \(x\): \[ \frac{d}{dx}[\cos(2x - 1)] = -\sin(2x - 1) \cdot 2 \] \[ \frac{d}{dx}[x^2] = 2x \] Now, we can substitute these derivatives back into the limit: \[ \lim_{x \to 0} \frac{-2\sin(2x - 1)}{2x} \] We can simplify this by canceling the 2s: \[ \lim_{x \to 0} \frac{-\sin(2x - 1)}{x} \] Now, we have another indeterminate form of \(\frac{0}{0}\), so we can apply L'Hôpital's Rule again: \[ \frac{d}{dx}[-\sin(2x - 1)] = -2\cos(2x - 1) \] \[ \frac{d}{dx}[x] = 1 \] Substituting these derivatives back into the limit: \[ \lim_{x \to 0} \frac{-2\cos(2x - 1)}{1} \] Now, we can evaluate this limit as \(x\) approaches 0: \[ \lim_{x \to 0} -2\cos(2x - 1) = -2\cos(-1) = -2\cos(1) \] Using the Maclaurin series expansion for \(\cos(x)\) around \(x = 0\), we know that: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \] So, for \(x = 1\): \[ \cos(1) = 1 - \frac{1}{2!} + \frac{1}{4!} - \cdots \] \[ \cos(1) \approx 1 - \frac{1}{2} + \frac{1}{24} - \cdots \] \[ \cos(1) \approx \frac{1}{2} + \frac{1}{24} - \cdots \] \[ \cos(1) \approx 0.540302305868139787328962 \] Therefore: \[ -2\cos(1) \approx -2 \times 0.540302305868139787328962 = -1.080604611736279574657924 \] So, the limit is approximately: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \approx -1.080604611736279574657924 \] This does not match the given solution of \(-2^2/2 = -2\). Therefore, the problem statement is incorrect. The correct limit for \(m = 2\) is approximately \(-1.0806\). |
True |
| doubao | First substitute m = 2 into the expression: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \] Now, using L'Hôpital's rule (since we have the indeterminate form 0/0), we take the derivative of the numerator and the denominator: The derivative of the numerator is $-2\sin(2x - 1)$ and the derivative of the denominator is $2x$. So the limit becomes: \[ \lim_{x \to 0} \frac{-2\sin(2x - 1)}{2x} = \lim_{x \to 0} -\sin(2x - 1)/x \] Applying L'Hôpital's rule again, the derivative of the numerator is $-2\cos(2x - 1)\times 2 = -4\cos(2x - 1)$ and the derivative of the denominator is 1. So the limit is: \[ \lim_{x \to 0} -4\cos(2x - 1) = -4\cos(-1) \neq -2 \] So the statement is not true for m = 2. |
True |
| minimax | To solve the given limit problem, we need to evaluate the limit as \( x \) approaches 0 for the function \( \frac{\cos(mx - 1)}{x^2} \) and check if it equals \( -\frac{m^2}{2} \) when \( m = 2 \). Let's start by plugging in \( m = 2 \) into the function: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \] To evaluate this limit, we can use L'Hôpital's Rule, which states that if the limit of functions \( f(x) \) and \( g(x) \) as \( x \) approaches a point \( c \) results in an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then the limit of \( \frac{f(x)}{g(x)} \) as \( x \) approaches \( c \) is the same as the limit of \( \frac{f'(x)}{g'(x)} \), provided the latter exists. First, we need to check if the original limit results in an indeterminate form. As \( x \) approaches 0, the numerator \( \cos(2x - 1) \) approaches \( \cos(-1) \), which is a finite number, and the denominator \( x^2 \) approaches 0. This gives us the indeterminate form \( \frac{\text{finite number}}{0} \), which is not directly in a form that L'Hôpital's Rule can be applied to. However, we can apply the rule to the numerator and denominator separately to see if it helps us evaluate the limit. Let's differentiate the numerator and the denominator with respect to \( x \): The derivative of the numerator \( \cos(2x - 1) \) with respect to \( x \) is \( -2\sin(2x - 1) \) (using the chain rule). The derivative of the denominator \( x^2 \) with respect to \( x \) is \( 2x \). Now we apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} = \lim_{x \to 0} \frac{-2\sin(2x - 1)}{2x} \] We can simplify the right-hand side by canceling out the 2's: \[ \lim_{x \to 0} \frac{-\sin(2x - 1)}{x} \] Now we apply L'Hôpital's Rule again because we still have an indeterminate form \( \frac{0}{0} \): Differentiate the numerator and the denominator again: The derivative of the numerator \( -\sin(2x - 1) \) with respect to \( x \) is \( -2\cos(2x - 1) \). The derivative of the denominator \( x \) with respect to \( x \) is \( 1 \). Applying L'Hôpital's Rule again: \[ \lim_{x \to 0} \frac{-2\cos(2x - 1)}{1} = -2\cos(-1) \] Since \( \cos(-1) \) is a finite number, the limit is simply \( -2 \) times that number. This does not match the given expression \( -\frac{m^2}{2} \) when \( m = 2 \), which would be \( -\frac{4}{2} = -2 \). Therefore, the limit does not equal \( -\frac{m^2}{2} \) when \( m = 2 \). There seems to be a mistake in the given problem statement, as the limit does not evaluate to \( -\frac{m^2}{2} \) for any value of \( m \), including \( m = 2 \). |
True |
| kimi | To find the limit of the given function as \( x \) approaches 0, 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 also exists and is equal to the limit of the ratio of their derivatives. The given function is: \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} \] For \( m = 2 \), the function becomes: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \] First, let's check if we can directly apply L'Hôpital's Rule. The limit as \( x \) approaches 0 of the numerator is: \[ \cos(2 \cdot 0) - 1 = \cos(0) - 1 = 1 - 1 = 0 \] And the limit of the denominator is: \[ 0^2 = 0 \] Since we have a form of \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We take the derivative of the numerator and the denominator with respect to \( x \): Numerator derivative: \[ \frac{d}{dx}(\cos(2x) - 1) = -2\sin(2x) \] Denominator derivative: \[ \frac{d}{dx}(x^2) = 2x \] Now we apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{-2\sin(2x)}{2x} \] This simplifies to: \[ \lim_{x \to 0} \frac{-2\sin(2x)}{2x} = \lim_{x \to 0} \frac{-\sin(2x)}{x} \] Now, we can use the fact that \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \) for any constant \( k \). In this case, \( k = 2 \), so: \[ \lim_{x \to 0} \frac{-\sin(2x)}{x} = -2 \] Therefore, the limit is: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] So, the statement \( \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \) is not true for \( m = 2 \). The correct limit is \( -2 \). |
True |
| wenxinyiyan | We can verify the given limit by evaluating it directly: \lim_{x \to 0} (\cos(mx - 1)/(x^2)) = \lim_{x \to 0} (\cos(2x - 1)/(x^2)) = \lim_{x \to 0} (1 - 2\sin^2(x - 1/2))/(x^2) = \lim_{x \to 0} (1 - 2(x - 1/2)^2)/(x^2) = \lim_{x \to 0} (1 - 2x^2 + 2x - 1/2)/(x^2) = \lim_{x \to 0} (-2x^2 + 2x - 1/2)/(x^2) = \lim_{x \to 0} (-2 + 2/x - 1/(2x^2)) As x approaches 0, the term 2/x approaches infinity and the term -1/(2x^2) approaches negative infinity, so the limit does not exist. Therefore, the statement that \lim_{x \to 0} (\cos(mx - 1)/(x^2)) = -(m^2)/2 for m = 2 is false. |
True |
| zhipu | To determine whether \(\lim_{x \to 0} \frac{\cos(mx - 1)}{x^2} = -\frac{m^2}{2}\) for \(m = 2\), we need to evaluate the limit and see if it matches the given value. First, let's substitute \(m = 2\) into the expression: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} \] To evaluate this limit, we can use L'Hôpital's Rule, which is applicable when we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Here, as \(x \to 0\), \(\cos(2x - 1) \to \cos(-1) \neq 0\) and \(x^2 \to 0\), so we have a \(\frac{\text{constant}}{0}\) form, which is not directly an indeterminate form. However, we can still apply L'Hôpital's Rule by considering the behavior of the numerator and denominator near \(x = 0\). First, let's find the derivatives of the numerator and the denominator: - The derivative of the numerator \(\cos(2x - 1)\) with respect to \(x\) is: \[ \frac{d}{dx}[\cos(2x - 1)] = -\sin(2x - 1) \cdot \frac{d}{dx}[2x - 1] = -2\sin(2x - 1) \] - The derivative of the denominator \(x^2\) with respect to \(x\) is: \[ \frac{d}{dx}[x^2] = 2x \] Now, applying L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\cos(2x - 1)}{x^2} = \lim_{x \to 0} \frac{-2\sin(2x - 1)}{2x} = \lim_{x \to 0} \frac{-\sin(2x - 1)}{x} \] We still have an indeterminate form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again: - The derivative of the numerator \(-\sin(2x - 1)\) with respect to \(x\) is: \[ \frac{d}{dx}[-\sin(2x - 1)] = -\cos(2x - 1) \cdot \frac{d}{dx}[2x - 1] = -2\cos(2x - 1) \] - The derivative of the denominator \(x\) with respect to \(x\) is: \[ \frac{d}{dx}[x] = 1 \] Applying L'Hôpital's Rule again: \[ \lim_{x \to 0} \frac{-\sin(2x - 1)}{x} = \lim_{x \to 0} \frac{-2\cos(2x - 1)}{1} = -2\cos(-1) \] Since \(\cos(-1) = \cos(1)\), we have: \[ -2\cos(1) \] Now, we need to check if this value equals \(-\frac{m^2}{2}\) for \(m = 2\): \[ -\frac{m^2}{2} = -\frac{2^2}{2} = -\frac{4}{2} = -2 \] Comparing \(-2\cos(1)\) with \(-2\): \[ -2\cos(1) \neq -2 \quad \text{(since } \cos(1) \neq 1 \text{)} \] Therefore, the limit \(\lim_{x \to 0} \frac{\cos(2x - 1)}{x^2}\) does not equal \(-\frac{m^2}{2}\) for \(m = 2\). ### Conclusion: The given statement \(\lim_{x \to 0} \frac{\cos(mx - 1)}{x^2} = -\frac{m^2}{2}\) for \(m = 2\) is **not true**. The actual limit evaluates to \(-2\cos(1)\), which is not equal to \(-2\). |
True |
| gpto1 | **Solution:** We are asked to determine whether the following limit holds true for \( m = 2 \): \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \] Specifically, we need to compute: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \] **Step 1: Understanding the Limit** The function involves the cosine function and a quadratic term in the denominator. As \( x \to 0 \), both the numerator and the denominator approach zero: - Numerator: \( \cos(2x) - 1 \to \cos(0) - 1 = 1 - 1 = 0 \) - Denominator: \( x^2 \to 0^2 = 0 \) Since we have an indeterminate form \( \frac{0}{0} \), we can use techniques like the Taylor series expansion or L'Hôpital's Rule to evaluate the limit. --- **Method 1: Using the Taylor Series Expansion** The Taylor series expansion of \( \cos(u) \) around \( u = 0 \) is: \[ \cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \frac{u^6}{720} + \cdots \] For very small \( x \), higher-order terms become negligible. Therefore, we can approximate \( \cos(2x) \) as: \[ \cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - \frac{4x^2}{2} = 1 - 2x^2 \] **Substituting into the limit:** \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \approx \lim_{x \to 0} \frac{(1 - 2x^2) - 1}{x^2} = \lim_{x \to 0} \frac{-2x^2}{x^2} = \lim_{x \to 0} -2 = -2 \] Thus, we find: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] Since \( m = 2 \), we compute \( -\frac{m^2}{2} \): \[ -\frac{m^2}{2} = -\frac{(2)^2}{2} = -\frac{4}{2} = -2 \] **Conclusion from Method 1:** The limit equals \( -2 \), which matches \( -\frac{m^2}{2} \) when \( m = 2 \). --- **Method 2: Using L'Hôpital's Rule** Since the limit is in the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to \( x \). **First Application of L'Hôpital's Rule:** Let: - Numerator \( N(x) = \cos(2x) - 1 \) - Denominator \( D(x) = x^2 \) Compute the derivatives: - \( N'(x) = -2\sin(2x) \) - \( D'(x) = 2x \) Apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = \lim_{x \to 0} \frac{N'(x)}{D'(x)} = \lim_{x \to 0} \frac{-2\sin(2x)}{2x} = \lim_{x \to 0} \frac{-\sin(2x)}{x} \] **Second Application of L'Hôpital's Rule:** Again, the limit is in the indeterminate form \( \frac{0}{0} \): - Numerator \( N_1(x) = -\sin(2x) \) - Denominator \( D_1(x) = x \) Compute the derivatives: - \( N_1'(x) = -2\cos(2x) \) - \( D_1'(x) = 1 \) Apply L'Hôpital's Rule again: \[ \lim_{x \to 0} \frac{-\sin(2x)}{x} = \lim_{x \to 0} \frac{N_1'(x)}{D_1'(x)} = \lim_{x \to 0} \frac{-2\cos(2x)}{1} = -2\cos(0) = -2(1) = -2 \] **Conclusion from Method 2:** Again, we find: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 = -\frac{(2)^2}{2} \] --- **Final Answer:** Yes, the limit holds true for \( m = 2 \): \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{(2)^2}{2} = -2 \] **Therefore,** \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \quad \text{when} \quad m = 2 \] |
True |
| deepseekv3 | To determine whether the limit \[ \lim_{x \to 0} \frac{\cos(mx) - 1}{x^2} = -\frac{m^2}{2} \] holds for \( m = 2 \), let's evaluate the limit step-by-step. ### Step 1: Substitute \( m = 2 \) We substitute \( m = 2 \) into the expression: \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} \] ### Step 2: Use the Taylor Series Expansion for \( \cos(2x) \) The Taylor series expansion for \( \cos(2x) \) around \( x = 0 \) is: \[ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \cdots \] Simplifying the first few terms: \[ \cos(2x) = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \cdots \] \[ \cos(2x) = 1 - 2x^2 + \frac{2x^4}{3} - \cdots \] ### Step 3: Substitute the Approximation into the Limit For small \( x \), higher-order terms (e.g., \( x^4 \)) become negligible. Thus, we approximate: \[ \cos(2x) - 1 \approx -2x^2 \] Substitute this into the limit: \[ \lim_{x \to 0} \frac{-2x^2}{x^2} \] ### Step 4: Simplify the Expression Cancel \( x^2 \) in the numerator and denominator: \[ \lim_{x \to 0} (-2) \] ### Step 5: Evaluate the Limit The limit is simply \(-2\): \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -2 \] ### Step 6: Compare with the Given Expression The given expression is: \[ -\frac{m^2}{2} \] Substitute \( m = 2 \): \[ -\frac{2^2}{2} = -\frac{4}{2} = -2 \] ### Final Answer Yes, the limit holds for \( m = 2 \): \[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{2^2}{2} = -2 \] |
True |