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Limits at 0 built from 1 - cos ax, including nested cosines

This type of question, asked 10 times: 4 May 2023 S1, 9 May 2024 S2, 10 May 2024 S2, 12 May 2024 S1, 25 Apr 2025 S1, 16 Apr 2026 S2, 19 Apr 2026 S1, 13 May 2026 S2, 14 May 2026 S2, 15 May 2026 S2. All 10 questions below, oldest first.

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  1. Question 1

    4 May 2023 S1

    147
    lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim_{x\to0}\frac{(1-\cos2x)(3+\cos x)}{x\tan4x} has the value

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  2. Question 2

    9 May 2024 S2

    140
    lim⁡x→0(1−cos⁡2x)⋅sin⁡5xx2sin⁡3x\lim_{x\to 0}\dfrac{(1-\cos 2x)\cdot\sin 5x}{x^2\sin 3x} is

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  3. Question 3

    10 May 2024 S2

    133
    If f(x)=4x4 ⁣[1−cos⁡x2−cos⁡x4+cos⁡x2⋅cos⁡x4]f(x)=\dfrac{4}{x^4}\!\left[1 - \cos\dfrac{x}{2} - \cos\dfrac{x}{4} + \cos\dfrac{x}{2}\cdot\cos\dfrac{x}{4}\right] is continuous at x=0x=0, then f(0)f(0) is

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  4. Question 4

    12 May 2024 S1

    144
    lim⁡x→0cos⁡7x∘−cos⁡2x∘x2\displaystyle\lim_{x\to0}\frac{\cos7x^{\circ}-\cos2x^{\circ}}{x^{2}} is

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  5. Question 5

    25 Apr 2025 S1

    133
    lim⁡x→52−2cos⁡(x2−12x+35)(x−5)=\lim_{x\rightarrow 5} \frac{\sqrt{2 - 2\cos\left( x^{2}- 12x+ 35 \right)}}{(x- 5)}=...... .

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  6. Question 6

    16 Apr 2026 S2

    70
    If the function f is continuous at x=πx = \pi, where f(x)=1−cos⁡[7(x−π)]5(x−π)2f(x) = \dfrac{1 - \cos[7(x - \pi)]}{5(x - \pi)^2}, for x≠πx \neq \pi, then f(π)=f(\pi) =

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  7. Question 7

    19 Apr 2026 S1

    61
    The value of lim⁡x→0(8x8)[1−cos⁡x22−cos⁡x24+cos⁡x22⋅cos⁡x24]\lim_{x \to 0}\left(\dfrac{8}{x^8}\right)\left[1 - \cos\dfrac{x^2}{2} - \cos\dfrac{x^2}{4} + \cos\dfrac{x^2}{2} \cdot \cos\dfrac{x^2}{4}\right] is equal to ...

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  8. Question 8

    13 May 2026 S2

    65
    lim⁡x→01−cos⁡(1−cos⁡x)x4=\lim_{x \to 0}\dfrac{1 - \cos(1 - \cos x)}{x^4} =

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  9. Question 9

    14 May 2026 S2

    72
    If f(x)=1−cos⁡(pπx2)x2f(x) = \dfrac{1 - \cos\left(\dfrac{p\pi x}{2}\right)}{x^2} is continuous at x=0x = 0 and f(0)=9π232f(0) = \dfrac{9\pi^2}{32}, then the value of pp is

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  10. Question 10

    15 May 2026 S2

    64
    Let f(x)f(x) be a function continuous at x=0x = 0, defined for x≠0x \ne 0 as f(x)=1−cos⁡(1−cos⁡(6x))x2sin⁡2(3x)f(x) = \dfrac{1 - \cos(1 - \cos(6x))}{x^2\sin^2(3x)}. Then the value of f(0)f(0) is...

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