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Product or quotient of standard limits (a^x - 1)/x, log(1 + x)/x, sin x/x, tan x/x

This type of question, asked 10 times: May 2021 S1, 9 May 2023 S1, 14 May 2024 S2, 20 Apr 2025 S1, 20 Apr 2025 S2, 13 Apr 2026 S1, 13 Apr 2026 S2, 15 Apr 2026 S1, 16 Apr 2026 S1, 20 Apr 2026 S1. All 10 questions below, oldest first.

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  1. Question 1

    May 2021 S1

    119
    If f(x)={sin⁡3(x)⋅log⁡(1+3x)(tan⁡−1x)2(e5x−1)xx≠0ax=0f(x) = \begin{cases} \dfrac{\sin^{3}(\sqrt{x}) \cdot \log(1 + 3x)}{\left(\tan^{-1}\sqrt{x}\right)^{2}\left(e^{5\sqrt{x}} - 1\right)x} & x \neq 0 \\ a & x = 0 \end{cases} is continuous in [0,1][0,1], then aa equals

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  2. Question 2

    9 May 2023 S1

    150
    Let k be a non-zero real number. If f(x)={(ex−1)2sin⁡(xk)log⁡(1+x4),x≠012,x=0f(x)=\begin{cases}\dfrac{(e^x-1)^2}{\sin\left(\frac{x}{k}\right)\log\left(1+\frac{x}{4}\right)}, & x\neq0 \\ 12, & x=0\end{cases} is a continuous function, then the value of k is

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  3. Question 3

    14 May 2024 S2

    150
    The function ff defined on [−13,13]\left[-\frac{1}{3},\frac{1}{3}\right] by f(x)={1xlog⁡1+3x1−2x,x≠0k,x=0f(x) = \begin{cases}\frac{1}{x}\log\frac{1+3x}{1-2x}, & x\neq0\\k, & x=0\end{cases} is continuous at x=0x=0, then kk is

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  4. Question 4

    20 Apr 2025 S1

    124
    lim⁡x→0etan⁡x−extan⁡x−x=\lim_{x\rightarrow 0} \frac{e^{\tan x}-e^{x}}{\tan x-x}=

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  5. Question 5

    20 Apr 2025 S2

    138
    lim⁡x→0(7x−1)4tan⁡(xk)⋅log⁡(1+x23)⋅sin⁡4x=3(log⁡7)3\lim_{x\rightarrow 0} \frac{\left( 7^{x}- 1 \right)^{4}}{\tan\left( \frac{x}{k} \right)\cdot \log\left( 1 +\frac{x^{2}}{3} \right)\cdot \sin4x}= 3(\log7)^{3}, then k=k =

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  6. Question 6

    13 Apr 2026 S1

    60
    lim⁡x→0[log⁡∣2+x∣−log⁡∣2−x∣tan⁡x]=…….\displaystyle\lim_{x \to 0}\left[\dfrac{\log|2 + x| - \log|2 - x|}{\tan x}\right] = \ldots\ldots.

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  7. Question 7

    13 Apr 2026 S2

    70
    If f(x)f(x) is continuous at x=0x = 0, where f(x)=8x−2xkx−1f(x) = \dfrac{8^x - 2^x}{k^x - 1}, for x≠0x \neq 0 and f(0)=2f(0) = 2, then the value of kk is ...

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  8. Question 8

    15 Apr 2026 S1

    61
    If lim⁡x→0(4x−1)3tan⁡(x4)log⁡(1+x23)=96(log⁡a)b\displaystyle\lim_{x \to 0}\dfrac{(4^x - 1)^3}{\tan\left(\dfrac{x}{4}\right)\log\left(1 + \dfrac{x^2}{3}\right)} = 96(\log a)^b, then (a+b)=(a + b) =

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  9. Question 9

    16 Apr 2026 S1

    60
    lim⁡x→0(5x−1)4 csc⁡ (xlog⁡5)tan⁡(xlog⁡5)⋅log⁡(1+x2log⁡25)=……\displaystyle\lim_{x \to 0}\dfrac{(5^x - 1)^4\,\mathrm{\csc}\,(x\log 5)}{\tan(x\log 5)\cdot\log(1 + x^2\log 25)} = \ldots\ldots

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  10. Question 10

    20 Apr 2026 S1

    62
    lim⁡x→0[x⋅log⁡(1+4x)(e4x−1)2]=⋯\lim_{x \to 0}\left[\dfrac{x \cdot \log(1 + 4x)}{(e^{4x} - 1)^2}\right] = \cdots

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