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7.5 I · Q64

Q.lim⁡x→π[5+cos⁡x−2(π−x)2]\displaystyle\lim_{x\to \pi}\left[\frac{\sqrt{5+\cos x}-2}{(\pi-x)^2}\right]

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Let t=π−xt=\pi-x, so x=π−tx=\pi-t and t→0t\to0. cos⁡x=−cos⁡t\cos x=-\cos t, so 5+cos⁡x=5−cos⁡t=4+(1−cos⁡t)=4+2sin⁡2(t/2)5+\cos x=5-\cos t=4+(1-\cos t)=4+2\sin^2(t/2). Rationalize the numerator by multiplying with 4+2sin⁡2(t/2)+2\sqrt{4+2\sin^2(t/2)}+2: [4+2sin⁡2(t/2)]−4=2sin⁡2(t/2)\left[4+2\sin^2(t/2)\right]-4=2\sin^2(t/2). So 5+cos⁡x−2=2sin⁡2(t/2)4+2sin⁡2(t/2)+2\sqrt{5+\cos x}-2=\dfrac{2\sin^2(t/2)}{\sqrt{4+2\sin^2(t/2)}+2}. Dividing by t2t^2 (the denominator): $\dfrac{2\sin^2(t/2)}{t^2\left[\sqrt{4+2\sin^ …

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