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Trigonometric limit as x tends to pi/2 with powers of (pi - 2x)

This type of question, asked 4 times: 9 May 2023 S2, 15 May 2023 S1, 9 May 2024 S1, 19 May 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    9 May 2023 S2

    131
    lim⁡x→π/2(1−sin⁡x)(8x3−π3)cos⁡x(π−2x)4\lim_{x\to\pi/2}\dfrac{(1-\sin x)(8x^3-\pi^3)\cos x}{(\pi-2x)^4} is equal to

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  2. Question 2

    15 May 2023 S1

    139
    lim⁡x→π/2(1−tan⁡x2)(1−sin⁡x)(1+tan⁡x2)(π−2x)3\lim_{x\to\pi/2}\frac{\left(1-\tan\frac{x}{2}\right)(1-\sin x)}{\left(1+\tan\frac{x}{2}\right)(\pi-2x)^{3}} is

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  3. Question 3

    9 May 2024 S1

    138
    lim⁡x→π/2cot⁡x−cos⁡x(π−2x)3\lim_{x\to\pi/2}\frac{\cot x-\cos x}{(\pi-2x)^3} equals

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  4. Question 4

    19 May 2026 S1

    64
    lim⁡x→π2[1−tan⁡(x2)](1−sin⁡x)[1+tan⁡(x2)](π−2x)3=\displaystyle\lim_{x \to \frac{\pi}{2}} \dfrac{[1 - \tan(\frac{x}{2})](1 - \sin x)}{[1 + \tan(\frac{x}{2})](\pi - 2x)^3} =

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