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Miscellaneous 7 I · Q109

Q.lim⁡x→π/2(3cos⁡x−1π2−x)=\displaystyle\lim_{x\to \pi/2}\left(\frac{3^{\cos x}-1}{\dfrac{\pi}{2}-x}\right)= (A) 11 (B) log⁡3\log 3 (C) 3π/23^{\pi/2} (D) 3log⁡33\log 3

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Let t=π/2−xt=\pi/2-x, so cos⁡x=cos⁡(π/2−t)=sin⁡t\cos x=\cos(\pi/2-t)=\sin t, and t→0t\to0 as x→π/2x\to\pi/2. The expression becomes 3sin⁡t−1t\dfrac{3^{\sin t}-1}{t}. As t→0t\to0, sin⁡t∼t\sin t\sim t, so $3^{\sin …

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