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Limit by rationalising (multiply by the conjugate)

This type of question, asked 6 times: 16 May 2023 S2, 10 May 2024 S1, 10 May 2024 S2, 13 May 2024 S2, 14 May 2024 S1, 13 May 2026 S1. All 6 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    16 May 2023 S2

    116
    lim⁡y→01+1+y4−2y4=\lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1+y^4}} - \sqrt{2}}{y^4} =

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  2. Question 2

    10 May 2024 S1

    150
    Value of lim⁡x→aa+2x−3x3a+x−2x\lim_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} is

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  3. Question 3

    10 May 2024 S2

    124
    lim⁡x→aa+2x−3x3a+x−2x=\displaystyle\lim_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}=

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  4. Question 4

    13 May 2024 S2

    116
    lim⁡x→01+xsin⁡x−cos⁡xtan⁡2(x/2)=\lim_{x\to0}\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{\tan^2(x/2)}=

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  5. Question 5

    14 May 2024 S1

    101
    lim⁡x→∞x3{x2+1+x4−x2}=\lim_{x\to\infty}x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x\sqrt2\right\}=

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  6. Question 6

    13 May 2026 S1

    63
    The value of lim⁡x→∞(x3+6x2+13−x)\lim_{x \to \infty}\left(\sqrt[3]{x^3 + 6x^2 + 1} - x\right) is:

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