| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the limit \(\lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x))\), we can analyze the behavior of the function as \(x\) approaches \(\frac{\pi}{2}\). First, note that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). As \(x\) approaches \(\frac{\pi}{2}\), \(\cos(x)\) approaches 0 and \(\sin(x)\) approaches 1. Therefore, \(\tan(x)\) approaches \(\infty\) (since the numerator approaches a non-zero value and the denominator approaches 0). The cosine function oscillates between -1 and 1. As its argument approaches \(\infty\), the function itself does not have a limit, as it keeps oscillating. However, for our limit, we are interested in the behavior of \(\cos(\tan(x))\) as \(x\) approaches \(\frac{\pi}{2}\). We know that \(\cos(x)\) approaches 0 as \(x\) approaches \(\frac{\pi}{2}\). Therefore, the product \(\cos(x)\cos(\tan(x))\) will be dominated by the behavior of \(\cos(x)\) as \(x\) approaches \(\frac{\pi}{2}\), because the other term, \(\cos(\tan(x))\), is bounded between -1 and 1 regardless of how quickly it oscillates. Hence, \(\lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x)) = \lim_{x \to \frac{\pi}{2}} \cos(x) \cdot \lim_{x \to \frac{\pi}{2}} \cos(\tan(x))\). We already established that \(\lim_{x \to \frac{\pi}{2}} \cos(x) = 0\). The limit of \(\cos(\tan(x))\) as \(x\) approaches \(\frac{\pi}{2}\) does not exist because it oscillates between -1 and 1, but since it is being multiplied by a term that approaches 0, the overall limit is 0. Therefore, the solution is: \[ \lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x)) = 0. \] |
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| deepseekr1 | To find the limit as \( x \) approaches \( \frac{\pi}{2} \) of \( \cos(x) \cos(\tan(x)) \): 1. **Behavior of \( \cos(x) \) near \( \frac{\pi}{2} \)**: - As \( x \) approaches \( \frac{\pi}{2} \), \( \cos(x) \) approaches 0. From the left, it is positive and approaching 0; from the right, it is negative and approaching 0. 2. **Behavior of \( \tan(x) \) near \( \frac{\pi}{2} \)**: - As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \tan(x) \) approaches \( +\infty \). From the right, it approaches \( -\infty \). 3. **Behavior of \( \cos(\tan(x)) \)**: - As \( \tan(x) \) approaches \( \pm\infty \), \( \cos(\tan(x)) \) oscillates between -1 and 1. However, it remains bounded within this interval. 4. **Product of \( \cos(x) \) and \( \cos(\tan(x)) \)**: - The function \( \cos(x) \) approaches 0, and \( \cos(\tan(x)) \) is bounded between -1 and 1. The product of a function approaching 0 and a bounded function will approach 0. 5. **Application of the Squeeze Theorem**: - Since \( |\cos(\tan(x))| \leq 1 \), we have \( |\cos(x) \cos(\tan(x))| \leq |\cos(x)| \). As \( x \) approaches \( \frac{\pi}{2} \), \( |\cos(x)| \) approaches 0. By the Squeeze Theorem, the limit of \( |\cos(x) \cos(\tan(x))| \) is 0, implying the original limit is also 0. Thus, the limit is \(\boxed{0}\). |
0 |
| gpto1 | To evaluate the limit: \[ \lim_{x \to \frac{\pi}{2}} \left( \cos(x) \cos\big(\tan(x)\big) \right), \] we need to consider the behavior of both \(\cos(x)\) and \(\cos\big(\tan(x)\big)\) as \(x\) approaches \(\frac{\pi}{2}\). **Step 1: Evaluate \(\cos(x)\) as \(x \to \frac{\pi}{2}\)** The cosine function \(\cos(x)\) is continuous everywhere, and specifically: \[ \cos\left( \frac{\pi}{2} \right) = 0. \] Additionally, near \(x = \frac{\pi}{2}\), we can use the approximation: \[ \cos\left( \frac{\pi}{2} - h \right) \approx \cos\left( \frac{\pi}{2} \right) + h \left( -\sin\left( \frac{\pi}{2} \right) \right) = 0 - h(1) = -h, \] where \(h = \frac{\pi}{2} - x\). So as \(x \to \frac{\pi}{2}\): \[ \cos(x) \approx -\left( \frac{\pi}{2} - x \right) \to 0^-. \] Thus, \(\cos(x) \to 0\). **Step 2: Evaluate \(\cos\big(\tan(x)\big)\) as \(x \to \frac{\pi}{2}\)** First, consider \(\tan(x)\) near \(x = \frac{\pi}{2}\): - As \(x \to \left( \frac{\pi}{2} \right)^-\), \(\tan(x) \to +\infty\). - As \(x \to \left( \frac{\pi}{2} \right)^+\), \(\tan(x) \to -\infty\). Since \(\tan(x)\) becomes unbounded, \(\cos\big(\tan(x)\big)\) will oscillate between \(-1\) and \(1\) because the cosine function is periodic and bounded: \[ -1 \leq \cos\big(\tan(x)\big) \leq 1. \] However, the exact value of \(\cos\big(\tan(x)\big)\) is not predictable due to the oscillation. Importantly, it remains bounded. **Step 3: Evaluate the Product** The product \( \cos(x) \cos\big(\tan(x)\big) \) involves a factor tending to \(0\) and another factor oscillating between \(-1\) and \(1\). Therefore: \[ \left| \cos(x) \cos\big(\tan(x)\big) \right| = \left| \cos(x) \right| \left| \cos\big(\tan(x)\big) \right| \leq \left| \cos(x) \right| \cdot 1. \] Since \(\cos(x) \to 0\), the entire expression tends to \(0\) regardless of the oscillation in \(\cos\big(\tan(x)\big)\). **Conclusion:** \[ \lim_{x \to \frac{\pi}{2}} \left( \cos(x) \cos\big(\tan(x)\big) \right) = 0. \] **Answer:** 0 |
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| GPT-4o | To find the limit of the function \(\lim_{x \to \frac{\pi}{2}} \cos(x) \cos(\tan(x))\), we need to carefully analyze the behavior of the cosine terms as \(x\) approaches \(\frac{\pi}{2}\). 1. **Behavior of \(\cos(x)\) near \(\frac{\pi}{2}\):** - As \(x\) approaches \(\frac{\pi}{2}\), \(\cos(x)\) approaches 0 because \(\cos(\frac{\pi}{2}) = 0\). 2. **Behavior of \(\tan(x)\) near \(\frac{\pi}{2}\):** - As \(x\) gets very close to \(\frac{\pi}{2}\), \(\tan(x)\) grows very large in magnitude (positive or negative infinity, depending on the direction of approach). 3. **Behavior of \(\cos(\tan(x))\) near \(\frac{\pi}{2}\):** - Since \(\tan(x)\) approaches \(\pm\infty\), we need to understand the behavior of \(\cos(\tan(x))\) as \(\tan(x)\) becomes very large. - The cosine function, \(\cos(y)\), oscillates between -1 and 1 for all real numbers \(y\). Therefore, even though \(\tan(x)\) becomes very large, \(\cos(\tan(x))\) will still oscillate between -1 and 1. 4. **Combining the two cosine terms:** - The product \(\cos(x) \cos(\tan(x))\) involves multiplying a term that approaches 0 (\(\cos(x)\)) by a term that stays bounded between -1 and 1 (\(\cos(\tan(x))\)). Given that \(\cos(x)\) approaches 0 as \(x\) approaches \(\frac{\pi}{2}\), and \(\cos(\tan(x))\) remains bounded, the product \(\cos(x) \cos(\tan(x))\) will approach 0. This is because any number multiplied by 0 is 0. Therefore, \[ \lim_{x \to \frac{\pi}{2}} \cos(x) \cos(\tan(x)) = 0. \] |
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| llama3.1-70B-Instruct | A nice trigonometric limit! Let's analyze the expression: $$\lim_{x \to \frac{\pi}{2}} (\cos(x) \cos(\tan(x)))$$ As $x$ approaches $\frac{\pi}{2}$, we know that: 1. $\cos(x)$ approaches $0$ (since $\cos(\frac{\pi}{2}) = 0$) 2. $\tan(x)$ approaches $\infty$ (since $\tan(\frac{\pi}{2})$ is undefined, but $\tan(x)$ increases without bound as $x$ approaches $\frac{\pi}{2}$) Now, let's consider the inner expression: $\cos(\tan(x))$. As $\tan(x)$ approaches $\infty$, we can use the fact that $\cos(u)$ approaches $0$ as $u$ approaches $\infty$ (since the cosine function oscillates between $-1$ and $1$ as its input increases without bound). Therefore, we can conclude that: $$\lim_{x \to \frac{\pi}{2}} \cos(\tan(x)) = 0$$ Now, we can use this result to evaluate the original limit: $$\lim_{x \to \frac{\pi}{2}} (\cos(x) \cos(\tan(x))) = \lim_{x \to \frac{\pi}{2}} (\cos(x) \cdot 0) = 0$$ So, the final answer is: $$\lim_{x \to \frac{\pi}{2}} (\cos(x) \cos(\tan(x))) = 0$$ |
0 |
| llama3.1-405B-Instruct | To evaluate the limit \lim_{x \to (\pi)/2} (cos(x)cos(tan(x))), we can use the following steps: 1. As x approaches π/2, cos(x) approaches 0. 2. Since tan(x) = sin(x)/cos(x), as x approaches π/2, tan(x) approaches infinity. 3. Since cos(tan(x)) is a periodic function with period π, as tan(x) approaches infinity, cos(tan(x)) approaches 0. 4. Therefore, as x approaches π/2, cos(x)cos(tan(x)) approaches 0. So, the limit \lim_{x \to (\pi)/2} (cos(x)cos(tan(x))) is 0. |
0 |
| DeepSeek-Coder-V2 | To solve the problem \(\lim_{x \to \frac{\pi}{2}} (\cos(x) \cos(\tan(x)))\), we need to analyze the behavior of the functions involved as \(x\) approaches \(\frac{\pi}{2}\). 1. **Analyze \(\cos(x)\) as \(x \to \frac{\pi}{2}\):** - The function \(\cos(x)\) approaches 0 as \(x\) approaches \(\frac{\pi}{2}\) because \(\cos(\frac{\pi}{2}) = 0\). 2. **Analyze \(\tan(x)\) as \(x \to \frac{\pi}{2}\):** - The function \(\tan(x)\) approaches \(\infty\) as \(x\) approaches \(\frac{\pi}{2}\) from the left (since \(\tan(x)\) is undefined at \(x = \frac{\pi}{2}\) but \(\lim_{x \to \frac{\pi}{2}^-} \tan(x) = \infty\)). 3. **Analyze \(\cos(\tan(x))\) as \(x \to \frac{\pi}{2}\):** - Since \(\tan(x) \to \infty\) as \(x \to \frac{\pi}{2}^-\), \(\cos(\tan(x))\) will approach \(\cos(\infty)\). The cosine function oscillates between -1 and 1 as its argument increases without bound, so \(\cos(\infty)\) does not have a specific limit but rather oscillates. 4. **Combine the results:** - We have \(\cos(x) \to 0\) and \(\cos(\tan(x))\) oscillates. Multiplying these two, we get \(\cos(x) \cos(\tan(x)) \to 0\) because the product of a function that approaches 0 and an oscillating function will still approach 0. Therefore, the limit is: \[ \lim_{x \to \frac{\pi}{2}} (\cos(x) \cos(\tan(x))) = 0 \] So, the final answer is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To find the limit \(\lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x))\), we can analyze the behavior of the function as \(x\) approaches \(\frac{\pi}{2}\). First, note that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). As \(x\) approaches \(\frac{\pi}{2}\), \(\cos(x)\) approaches 0 and \(\sin(x)\) approaches 1. Therefore, \(\tan(x)\) approaches \(\infty\) (since the numerator approaches a non-zero value and the denominator approaches 0). The cosine function oscillates between -1 and 1. As its argument approaches \(\infty\), the function itself does not have a limit, as it keeps oscillating. However, for our limit, we are interested in the behavior of \(\cos(\tan(x))\) as \(x\) approaches \(\frac{\pi}{2}\). We know that \(\cos(x)\) approaches 0 as \(x\) approaches \(\frac{\pi}{2}\). Therefore, the product \(\cos(x)\cos(\tan(x))\) will be dominated by the behavior of \(\cos(x)\) as \(x\) approaches \(\frac{\pi}{2}\), because the other term, \(\cos(\tan(x))\), is bounded between -1 and 1 regardless of how quickly it oscillates. Hence, \(\lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x)) = \lim_{x \to \frac{\pi}{2}} \cos(x) \cdot \lim_{x \to \frac{\pi}{2}} \cos(\tan(x))\). We already established that \(\lim_{x \to \frac{\pi}{2}} \cos(x) = 0\). The limit of \(\cos(\tan(x))\) as \(x\) approaches \(\frac{\pi}{2}\) does not exist because it oscillates between -1 and 1, but since it is being multiplied by a term that approaches 0, the overall limit is 0. Therefore, the solution is: \[ \lim_{x \to \frac{\pi}{2}} \cos(x)\cos(\tan(x)) = 0. \] |
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| baichuan4 | To find the limit of cos(x)cos(tan(x)) as x approaches π/2, we can use the following trigonometric identity: cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)] Applying this identity to our limit problem, we get: \lim_{x \to \pi/2} (cos(x)cos(tan(x))) = \lim_{x \to \pi/2} (1/2)[cos(x + tan(x)) + cos(x - tan(x))] Now, we need to evaluate the limits of cos(x + tan(x)) and cos(x - tan(x)) as x approaches π/2. First, let's consider cos(x + tan(x)) as x approaches π/2: x + tan(x) = π/2 + tan(π/2) Since tan(π/2) is undefined, we cannot directly evaluate this limit. However, we can consider the behavior of tan(x) as x approaches π/2 from below (left-hand limit) and from above (right-hand limit). As x approaches π/2 from below (0 < x < π/2), tan(x) approaches positive infinity, so x + tan(x) approaches π/2 + ∞. As x approaches π/2 from above (π/2 < x < π), tan(x) approaches negative infinity, so x + tan(x) approaches π/2 - ∞. In both cases, cos(x + tan(x)) will approach 0, since cos(∞) = 0 and cos(-∞) = 0. Next, let's consider cos(x - tan(x)) as x approaches π/2: x - tan(x) = π/2 - tan(π/2) Again, since tan(π/2) is undefined, we cannot directly evaluate this limit. However, we can consider the behavior of tan(x) as x approaches π/2 from below and from above. As x approaches π/2 from below (0 < x < π/2), tan(x) approaches positive infinity, so x - tan(x) approaches π/2 - ∞. As x approaches π/2 from above (π/2 < x < π), tan(x) approaches negative infinity, so x - tan(x) approaches π/2 + ∞. In both cases, cos(x - tan(x)) will approach 0, since cos(∞) = 0 and cos(-∞) = 0. Therefore, we have: \lim_{x \to \pi/2} (cos(x)cos(tan(x))) = \lim_{x \to \pi/2} (1/2)[0 + 0] = 0 So, the limit of cos(x)cos(tan(x)) as x approaches π/2 is 0. |
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| doubao | As x approaches π/2, cos(x) approaches 0 and tan(x) approaches infinity. But cos(tan(x)) oscillates between -1 and 1. The product of a number approaching 0 and a bounded oscillating quantity is 0. So, the limit is 0. | 0 |
| minimax | To find the limit \(\lim_{x \to (\pi)/2} \cos(x)\cos(\tan(x))\), we need to consider the behavior of the trigonometric functions involved as \(x\) approaches \(\pi/2\). As \(x\) approaches \(\pi/2\), \(\cos(x)\) approaches 0 because \(\cos(\pi/2) = 0\). The function \(\tan(x)\) approaches infinity as \(x\) approaches \(\pi/2\) from the left (since \(\tan(x) = \sin(x)/\cos(x)\), and \(\sin(\pi/2) = 1\) while \(\cos(\pi/2) = 0\), making the fraction undefined at \(\pi/2\)). However, the cosine function is continuous everywhere, and since \(\cos(\tan(x))\) is composed of continuous functions (the tangent function and the cosine function), it is continuous everywhere as well. But as \(x\) approaches \(\pi/2\), \(\tan(x)\) becomes very large, and the value of \(\cos(\tan(x))\) will oscillate between -1 and 1 rapidly due to the periodic nature of the cosine function. The product of a number approaching zero and a number oscillating between -1 and 1 without settling on a particular value is indeterminate. However, in this case, we can use the squeeze theorem to determine the limit. The squeeze theorem states that if for all \(x\) in some interval containing \(c\) (except possibly at \(c\) itself), \(g(x) \leq f(x) \leq h(x)\), and \(\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} f(x) = L\). In our case, we can't directly apply the squeeze theorem because \(\cos(\tan(x))\) doesn't have a limit as \(x\) approaches \(\pi/2\). However, we can say that the product \(\cos(x)\cos(\tan(x))\) will approach 0 because \(\cos(x)\) approaches 0 and the other factor, \(\cos(\tan(x))\), is bounded between -1 and 1. Therefore, the limit of the product as \(x\) approaches \(\pi/2\) is: \[ \lim_{x \to (\pi)/2} \cos(x)\cos(\tan(x)) = 0 \] This is because the product of a number approaching zero and a bounded function will always approach zero. |
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| kimi | To solve the limit \(\lim_{x \to (\pi/2)} (\cos(x)\cos(\tan(x)))\), we need to evaluate the behavior of the function as \(x\) approaches \(\pi/2\). Let's break it down step by step. 1. **Evaluate \(\cos(x)\) as \(x \to \pi/2\)**: \[ \cos(\pi/2) = 0 \] So, \(\cos(x)\) approaches 0 as \(x\) approaches \(\pi/2\). 2. **Evaluate \(\tan(x)\) as \(x \to \pi/2\)**: \[ \tan(\pi/2) \text{ is undefined} \] However, we are interested in the limit as \(x\) approaches \(\pi/2\) from the left (since \(\tan(x)\) is undefined at \(\pi/2\) from the right). As \(x\) approaches \(\pi/2\) from the left, \(\tan(x)\) approaches \(-\infty\). 3. **Evaluate \(\cos(\tan(x))\) as \(x \to \pi/2\)**: Since \(\tan(x)\) approaches \(-\infty\) as \(x\) approaches \(\pi/2\), \(\cos(\tan(x))\) is the cosine of a very large negative number. As the cosine function oscillates between -1 and 1, the limit of \(\cos(\tan(x))\) as \(\tan(x) \to -\infty\) does not exist in the traditional sense because it does not approach a single value. 4. **Combine the results**: We have \(\cos(x) \to 0\) as \(x \to \pi/2\) and \(\cos(\tan(x))\) does not approach a single value. However, since \(\cos(x)\) is approaching 0, the product \(\cos(x)\cos(\tan(x))\) will approach 0, because multiplying any value by 0 gives 0. Therefore, the limit is: \[ \lim_{x \to (\pi/2)} (\cos(x)\cos(\tan(x))) = 0 \] |
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| wenxinyiyan | 我们可以利用三角函数的性质和极限的运算法则来解决这个问题。 首先，我们注意到当$x \to \frac{\pi}{2}$时，$\tan(x) \to \infty$。因此，我们需要找到一个方法来处理$\cos(\tan(x))$这一项。 我们可以将$\cos(\tan(x))$写成$\frac{1}{\sqrt{1+\tan^2(x)}}$，这样当$\tan(x) \to \infty$时，这一项就趋于0。 因此，我们有： $\lim_{x \to \frac{\pi}{2}} (\cos(x)\cos(\tan(x))) = \lim_{x \to \frac{\pi}{2}} \left( \cos(x) \cdot \frac{1}{\sqrt{1+\tan^2(x)}} \right)$ 由于$\cos(x) \to 0$，而$\frac{1}{\sqrt{1+\tan^2(x)}} \to 0$，所以整个极限趋于0。 因此，答案是0。 |
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| zhipu | To find the limit \(\lim_{x \to \frac{\pi}{2}} \cos(x) \cos(\tan(x))\), we need to analyze the behavior of the function as \(x\) approaches \(\frac{\pi}{2}\). First, let's consider the individual components of the function: 1. **\(\cos(x)\) as \(x \to \frac{\pi}{2}\):** \[ \cos\left(\frac{\pi}{2}\right) = 0 \] So, as \(x\) approaches \(\frac{\pi}{2}\), \(\cos(x)\) approaches 0. 2. **\(\tan(x)\) as \(x \to \frac{\pi}{2}\):** \[ \tan(x) \to \infty \quad \text{as} \quad x \to \frac{\pi}{2} \] This is because the tangent function approaches infinity as its argument approaches \(\frac{\pi}{2}\) from either side. 3. **\(\cos(\tan(x))\) as \(x \to \frac{\pi}{2}\):** Since \(\tan(x)\) approaches infinity, we need to consider the behavior of \(\cos(y)\) as \(y \to \infty\). The cosine function oscillates between -1 and 1 for all real \(y\), and thus \(\cos(\tan(x))\) will also oscillate between -1 and 1 regardless of how large \(\tan(x)\) becomes. Now, combining these observations: - \(\cos(x) \to 0\) as \(x \to \frac{\pi}{2}\). - \(\cos(\tan(x))\) oscillates between -1 and 1. The product \(\cos(x) \cos(\tan(x))\) will be the product of a number approaching 0 and a number that oscillates between -1 and 1. Regardless of the oscillations of \(\cos(\tan(x))\), the fact that \(\cos(x)\) is approaching 0 will dominate the behavior of the product. Therefore, the limit of the product as \(x\) approaches \(\frac{\pi}{2}\) is: \[ \lim_{x \to \frac{\pi}{2}} \cos(x) \cos(\tan(x)) = 0 \] So, the solution to the problem is: \[ \boxed{0} \] |
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| gpto1 | To evaluate the limit: \[ \lim_{x \to \frac{\pi}{2}} \left( \cos(x) \cos\big(\tan(x)\big) \right), \] we need to consider the behavior of both \(\cos(x)\) and \(\cos\big(\tan(x)\big)\) as \(x\) approaches \(\frac{\pi}{2}\). **Step 1: Evaluate \(\cos(x)\) as \(x \to \frac{\pi}{2}\)** The cosine function \(\cos(x)\) is continuous everywhere, and specifically: \[ \cos\left( \frac{\pi}{2} \right) = 0. \] Additionally, near \(x = \frac{\pi}{2}\), we can use the approximation: \[ \cos\left( \frac{\pi}{2} - h \right) \approx \cos\left( \frac{\pi}{2} \right) + h \left( -\sin\left( \frac{\pi}{2} \right) \right) = 0 - h(1) = -h, \] where \(h = \frac{\pi}{2} - x\). So as \(x \to \frac{\pi}{2}\): \[ \cos(x) \approx -\left( \frac{\pi}{2} - x \right) \to 0^-. \] Thus, \(\cos(x) \to 0\). **Step 2: Evaluate \(\cos\big(\tan(x)\big)\) as \(x \to \frac{\pi}{2}\)** First, consider \(\tan(x)\) near \(x = \frac{\pi}{2}\): - As \(x \to \left( \frac{\pi}{2} \right)^-\), \(\tan(x) \to +\infty\). - As \(x \to \left( \frac{\pi}{2} \right)^+\), \(\tan(x) \to -\infty\). Since \(\tan(x)\) becomes unbounded, \(\cos\big(\tan(x)\big)\) will oscillate between \(-1\) and \(1\) because the cosine function is periodic and bounded: \[ -1 \leq \cos\big(\tan(x)\big) \leq 1. \] However, the exact value of \(\cos\big(\tan(x)\big)\) is not predictable due to the oscillation. Importantly, it remains bounded. **Step 3: Evaluate the Product** The product \( \cos(x) \cos\big(\tan(x)\big) \) involves a factor tending to \(0\) and another factor oscillating between \(-1\) and \(1\). Therefore: \[ \left| \cos(x) \cos\big(\tan(x)\big) \right| = \left| \cos(x) \right| \left| \cos\big(\tan(x)\big) \right| \leq \left| \cos(x) \right| \cdot 1. \] Since \(\cos(x) \to 0\), the entire expression tends to \(0\) regardless of the oscillation in \(\cos\big(\tan(x)\big)\). **Conclusion:** \[ \lim_{x \to \frac{\pi}{2}} \left( \cos(x) \cos\big(\tan(x)\big) \right) = 0. \] **Answer:** 0 |
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