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Find constants that make a limit finite, by series expansion

This type of question, asked 11 times: 2021 Paper 4, 2021 Paper 5, 26 Jul 2022, 29 Jul 2022, 13 Apr 2023, 27 Jan 2024, 31 Jan 2024, 5 Apr 2024, 2 Apr 2025, 4 Apr 2025, 22 Jan 2026 S2. All 12 questions below, oldest first.

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  1. Question 1

    2021 Paper 4

    81
    If lim⁡x→0ax−(e4x−1)ax(e4x−1)\lim_{x\rightarrow 0} \frac{ax-\left( e^{4x}- 1 \right)}{ax\left( e^{4x}- 1 \right)} exists and is equal to bb, then the value of a−2ba- 2b is

  2. Question 2

    2021 Paper 5

    84
    If lim⁡x→0aex−bcosx+ce−xxsinx=2\lim_{x \rightarrow 0} \frac{ae^{x}-bcosx+ce^{-x}}{xsinx}= 2, then a+b+ca+b+c is equal to

  3. Question 3

    26 Jul 2022

    157
    Let β=lim⁡x→0αx−(e3x−1)αx(e3x−1)\beta =\lim_{x \rightarrow 0} \frac{\alpha x -\left( e^{3x}- 1 \right)}{\alpha x\left( e^{3x}- 1 \right)} for some α∈R\alpha \in R. Then the value of α+β\alpha + \beta is:

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  4. Question 4

    29 Jul 2022

    65
    If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim_{x\rightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x\sin^{2}x}=\frac{2}{3}, where α,β,γ∈R\alpha,\beta,\gamma\in R, then which of the following is NOT correct?

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  5. Question 5

    13 Apr 2023

    159
    If lim⁡x→0eax−cos⁡(bx)−cxe−cx21−cos⁡(2x)=17\lim_{x \rightarrow 0} \frac{e^{ax}- \cos(bx) -\frac{cxe^{- cx}}{2}}{1 - \cos(2x)}= 17, then 5a2+b25a^{2}+b^{2} is equal to

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  6. Question 6

    27 Jan 2024

    159
    If lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13\lim_{x \rightarrow 0} \frac{3 + \alpha \sin x + \beta \cos x +\log_{e}(1 - x)}{3\tan^{2}x}=\frac{1}{3}, then 2α−β2\alpha - \beta is equal to :

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  7. Question 7

    31 Jan 2024

    177
    If lim⁡x→0ax2ex−blog⁡e(1+x)+cxe−xx2sin⁡x=1\lim_{x\rightarrow 0} \frac{ax^{2}e^{x}-b\log_{e}(1 +x) +cxe^{-x}}{x^{2}\sin x}= 1, then 16(a2+b2+c2)16\left( a^{2}+b^{2}+c^{2} \right) is equal to

  8. Question 8

    5 Apr 2024

    68
    If the function f(x)=sin⁡3x+αsin⁡x−βcos⁡3xx3f(x) =\frac{\sin3x + \alpha \sin x - \beta \cos3x}{x^{3}}, x∈Rx \in R, is continuous at x=0x = 0, then f(0)f(0) is equal to :

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  9. Question 9

    2 Apr 2025

    139
    If lim⁡x→0cos⁡(2x)+acos(4x)−bx4\lim_{x\rightarrow 0} \frac{\cos(2x) +acos(4x) -b}{x^{4}} is finite, then (a+b)(a+b) is equal to:

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  10. Question 10

    2 Apr 2025

    63
    For α,β,γ∈R\alpha,\beta,\gamma \in R, if lim⁡x→0x2sin⁡αx+(γ−1)ex2sin⁡2x−βx=3\lim_{x \rightarrow 0} \frac{x^{2}\sin\alpha x + (\gamma- 1)e^{x^{2}}}{\sin2x -\beta x}= 3, then β+γ−α\beta+\gamma-\alpha is equal to:

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  11. Question 11

    4 Apr 2025

    57
    If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x\rightarrow1^{+}} \frac{(x- 1)(6 +\lambda \cos(x- 1)) +\mu \sin(1 -x)}{(x- 1)^{3}}= - 1, where λ,μ∈R\lambda,\mu\in\mathbb{R}, then λ+μ\lambda+\mu is equal to.

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  12. Question 12

    22 Jan 2026 S2

    55
    If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim_{x \rightarrow 0} \frac{e^{(a - 1)x}+ 2\cos bx + (c - 2)e^{- x}}{x\cos x -\log_{e}(1 + x)}= 2, then a2+b2+c2a^{2}+b^{2}+c^{2} is equal to :

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