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(1 - cos 4x)/x^2 on the left, root(x)/(root(16 + root(x)) - 4) on the right: find the value at 0

The same question, asked 3 times: 11 May 2023 S2, 12 May 2024 S2, 22 Apr 2025 S2. Every wording below, oldest first.

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  1. Question 1

    11 May 2023 S2

    131
    Let f(x)={1−cos⁡4xx2,x<0a,x=0x16+x−4,x>0f(x) = \begin{cases}\frac{1-\cos 4x}{x^2}, & x<0\\ a, & x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{cases}. If f(x)f(x) is continuous at x=0x=0, then the value of a is

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  2. Question 2

    12 May 2024 S2

    120
    Given f(x)={1−cos⁡4xx2,x<0α,x=0x16+x−4,x>0f(x)=\begin{cases}\dfrac{1-\cos4x}{x^2}, & x<0\\ \alpha, & x=0\\ \dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{cases}. If f(x)f(x) is continuous at x=0x=0, then value of α\alpha is

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  3. Question 3

    22 Apr 2025 S2

    137
    If f(x)={1−cos⁡4xx2,x<0a,x=0x(16+x)1/2−4,x>0f(x)=\begin{cases}\dfrac{1-\cos4x}{x^2}, & x<0\\ a, & x=0\\ \dfrac{\sqrt{x}}{(16+\sqrt{x})^{1/2}-4}, & x>0\end{cases} is continuous at x=0x=0, then a=a=

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