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lim(n to oo) x^(2)sin(log(e)sqrt(cos((pi...

`lim_(n to oo) x^(2)sin(log_(e)sqrt(cos((pi)/(x))))` is less then

A

`0`

B

`-(pi^(2))/(2)`

C

`-(pi^(2))/(4)`

D

`(pi^(2))/(8)`

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The correct Answer is:
To solve the limit \( \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \), we will follow these steps: ### Step 1: Simplify the expression We start with the limit: \[ \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \] We can rewrite the sine function using the logarithmic identity: \[ \sin(a) \approx a \quad \text{for small } a \] Thus, we can approximate: \[ \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \approx \log_e\sqrt{\cos\left(\frac{\pi}{x}\right)} \] ### Step 2: Rewrite the logarithm Using the property of logarithms: \[ \log_e\sqrt{y} = \frac{1}{2} \log_e y \] we can express our limit as: \[ \lim_{x \to \infty} x^2 \cdot \frac{1}{2} \log_e\left(\cos\left(\frac{\pi}{x}\right)\right) \] ### Step 3: Analyze \(\cos\left(\frac{\pi}{x}\right)\) As \(x \to \infty\), \(\frac{\pi}{x} \to 0\). Therefore, we can use the Taylor expansion for \(\cos\): \[ \cos\left(\frac{\pi}{x}\right) \approx 1 - \frac{1}{2}\left(\frac{\pi}{x}\right)^2 \] Thus, we have: \[ \log_e\left(\cos\left(\frac{\pi}{x}\right)\right) \approx \log_e\left(1 - \frac{1}{2}\left(\frac{\pi}{x}\right)^2\right) \] ### Step 4: Use the logarithmic approximation Using the approximation \(\log_e(1 + u) \approx u\) for small \(u\): \[ \log_e\left(1 - \frac{1}{2}\left(\frac{\pi}{x}\right)^2\right) \approx -\frac{1}{2}\left(\frac{\pi}{x}\right)^2 \] So, we can substitute this back into our limit: \[ \lim_{x \to \infty} x^2 \cdot \frac{1}{2} \left(-\frac{1}{2}\left(\frac{\pi}{x}\right)^2\right) \] ### Step 5: Simplify the limit This simplifies to: \[ \lim_{x \to \infty} x^2 \cdot \frac{-\pi^2}{4x^2} = \lim_{x \to \infty} \frac{-\pi^2}{4} = -\frac{\pi^2}{4} \] ### Final Result Thus, the limit is: \[ \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) = -\frac{\pi^2}{4} \] ### Conclusion The answer is: \[ \text{Option 3: } -\frac{\pi^2}{4} \]

To solve the limit \( \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \), we will follow these steps: ### Step 1: Simplify the expression We start with the limit: \[ \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \] We can rewrite the sine function using the logarithmic identity: ...
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