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Limits and Continuity

JEE Mains Maths. Try each one first, then open its answer.

Top 10 JEE Mains PYQs to practise first

The questions and kinds of question asked most often come first. Each one says how often and in which papers.

  1. Question 1

    This type asked 16 times: 2021 Paper 10, 2021 Paper 14, 2021 Paper 18, 2021 Paper 22, 2021 Paper 26, 26 Jul 2022, 29 Jul 2022, 27 Jan 2024, 31 Jan 2024, 4 Apr 2024, 23 Jan 2025, 3 Apr 2025, 23 Jan 2026 S1, 23 Jan 2026 S2, 24 Jan 2026 S2, 8 Apr 2026 S2. This one: 8 Apr 2026 S2 See all 16 of this type

    69
    Let f(x)={13,x≤π/2b(1−sin⁡x)(π−2x)2,x>π/2f(x) =\left\{ \begin{matrix} \frac{1}{3} & , & x\leq\pi/2 \\ \frac{b(1 - \sin x)}{(\pi- 2x)^{2}} & , & x>\pi/2 \end{matrix} \right., If ff is continuous at x=π/2x =\pi/2, then the value of ∫03 b−6∣x2+2x−3∣dx\int_{0}^{3\text{ }b - 6} \left| x^{2}+ 2x - 3 \right|dx is :

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  2. Question 2

    This type 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. This one: 22 Jan 2026 S2 See all 11 of this type

    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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  3. Question 3

    This type asked 8 times: 2021 Paper 3, 28 Jul 2022, 22 Jan 2025, 23 Jan 2025, 29 Jan 2025, 3 Apr 2025, 4 Apr 2025, 21 Jan 2026 S1. This one: 21 Jan 2026 S1 See all 8 of this type

    60
    Let f:R→(0,∞)f:R \rightarrow (0,\infty) be a twice differentiable function such that f(3)=18,f′(3)=0f(3) = 18,f^{'}(3) = 0 and f′′(3)=4f^{''}(3) = 4. Then lim⁡x→1(log⁡e(f(2+x)f(3))18(x−1)2)\lim_{x \rightarrow 1} \left( \log_{e}\left( \frac{f(2 + x)}{f(3)} \right)^{\frac{18}{(x - 1)^{2}}} \right) is equal to :

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  4. Question 4

    This type asked 7 times: 14 Jun 2022, 12 Apr 2023, 5 Apr 2024, 6 Apr 2024, 7 Apr 2025, 22 Jan 2026 S2, 2 Apr 2026 S2. This one: 2 Apr 2026 S2 See all 7 of this type

    75
    The number of points in the interval [2,4], at which the function f(x)=[x2−x−12]f(x) = \left\lbrack x^{2} - x - \frac{1}{2} \right\rbrack, where [⋅]\lbrack \cdot \rbrack denotes the greatest integer function, is discontinuous, is ____\_\_\_\_ .

  5. Question 5

    This type asked 6 times: 2021 Paper 1, 2021 Paper 24, 30 Jan 2023, 29 Jan 2024, 30 Jan 2024, 4 Apr 2024. This one: 4 Apr 2024 See all 6 of this type

    158
    Let f(x)=∫0x(t+sin⁡(1−et))dt,x∈Rf(x) =\int_{0}^{x} \left( t + \sin\left( 1 -e^{t} \right) \right)dt,x \in R. Then lim⁡x→0f(x)x3\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}} is equal to

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  6. Question 6

    This type asked 6 times: 2021 Paper 4, 2021 Paper 12, 2021 Paper 18, 2021 Paper 20, 25 Jul 2022, 1 Feb 2023. This one: 1 Feb 2023 See all 6 of this type

    73
    lim⁡n→∞[11+n+12+n+13+n+⋯+12n]\lim_{n \rightarrow \infty} \left\lbrack \frac{1}{1 + n}+\frac{1}{2 + n}+\frac{1}{3 + n}+ \cdots +\frac{1}{2n} \right\rbrack is equal to

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  7. Question 7

    This type asked 5 times: 2021 Paper 6, 25 Jun 2022, 28 Jun 2022, 26 Jul 2022, 6 Apr 2026 S2. This one: 6 Apr 2026 S2 See all 5 of this type

    75
    Let f(x)={x3+8x<0x2−4x≥0f(x) = \left\{ \begin{matrix} x^{3} + 8 & x < 0 \\ x^{2} - 4 & x \geq 0 \end{matrix} \right. and g(x)={(x−8)1/3;x<0(x+4)1/2;x≥0g(x) = \left\{ \begin{matrix} (x - 8)^{1/3}; & x < 0 \\ (x + 4)^{1/2}; & x \geq 0 \end{matrix} \right. Then the number of points, where the function gof is discontinuous, is ____\_\_\_\_ .

  8. Question 8

    This type asked 3 times: 8 Apr 2024, 28 Jan 2026 S1, 5 Apr 2026 S1. This one: 5 Apr 2026 S1 See all 3 of this type

    67
    The product of all possible values of α\alpha, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2\lim_{x\rightarrow 0} \left( \frac{1 - \cos(\alpha x)\cos((\alpha+ 1)x)\cos((\alpha+ 2)x)}{\sin^{2}((\alpha+ 1)x)} \right)= 2, is :

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  9. Question 9

    This type asked 3 times: 2021 Paper 16, 25 Jan 2023, 9 Apr 2024. This one: 9 Apr 2024 See all 3 of this type

    86
    Let f:(0,π)→Rf:(0,\pi) \rightarrow R be a function given by
    f(x)={(87)tan⁡8xtan⁡7x,0<x<π2a−8,x=π2(1+∣cot⁡x∣)ba∣tan⁡x∣,π2<x<πf(x) =\left\{ \begin{matrix} \left( \frac{8}{7} \right)^{\frac{\tan8x}{\tan7x}}, & 0 <x<\frac{\pi}{2} \\ a- 8, & x=\frac{\pi}{2} \\ (1 + |\cot x|)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2}<x<\pi \end{matrix} \right.
    Where a,b∈Za,b \in Z. If ff is continuous at x=π2x =\frac{\pi}{2}, then a2+b2a^{2}+b^{2} is equal to

  10. Question 10

    This type asked 3 times: 2021 Paper 20, 8 Apr 2023, 5 Apr 2024. This one: 5 Apr 2024 See all 3 of this type

    176
    Let a>0a > 0 be a root of the equation 2x2+x−2=02x^{2}+ x - 2 = 0. If lim⁡x→1a16(1−cos⁡(2+x−2x2))(1−ax)2=α+β17\lim_{x \rightarrow\frac{1}{a}} \frac{16\left( 1 - \cos\left( 2 + x - 2x^{2} \right) \right)}{\left( 1 - ax \right)^{2}}= \alpha + \beta\sqrt{17}, where α,β∈Z\alpha,\beta \in Z then α+β\alpha + \beta is equal to