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Maths · chapter-wise PYQs

Limits: JEE Main 2026 PYQs

Every 2026 Session 2 question on this chapter, in one place.

Last updated: 26 October 2026

Direct answer

JEE Main 2026 asked 5 questions on Limits across 5 shifts (Session 2).

Revise the chapter first: Limits notes and formulas. Then solve these without looking at solutions.

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Limits questions

Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.

2 April 2026 Shift 1, Q16 · Q16 · MCQ · Limits and Derivatives · Hard

If $\lim_{x\to 2} \frac{\sin(x^3-5x^2+ax+b)}{(\sqrt{x-1}-1)\log_e(x-1)} = m$, then $a+b+m$ is equal to:

  1. (1) 5
  2. (2) 6
  3. (3) 8
  4. (4) 10
Show answer and solution

Answer: Option (2)

Key idea: Limits with indeterminate forms, Taylor series expansion, and parameter determination

  1. For the limit to exist, both numerator and denominator must approach 0 as $x \to 2$. This gives $2a + b = 12$.
  2. Substitute $u = x - 2$ and expand $x^3 - 5x^2 + ax + b = (a-8)u + u^2 + u^3 + (2a+b-12)$.
  3. For a non-zero finite limit, the coefficient of $u$ must be zero: $a - 8 = 0$, so $a = 8$ and $b = -4$.
  4. Using Taylor expansions: $\sqrt{x-1} - 1 \approx \frac{u}{2}$ and $\ln(x-1) \approx u$ as $u \to 0$.
  5. The limit becomes $\lim_{u \to 0} \frac{\sin(u^2(1+u))}{\frac{u^2}{2}} = \lim_{u \to 0} \frac{u^2(1+u)}{\frac{u^2}{2}} = 2$, so $m = 2$.
  6. Therefore, $a + b + m = 8 + (-4) + 2 = 6$.

5 April 2026 Shift 1, Q17 · Q17 · MCQ · Limits and Derivatives · Hard

The product of all possible values of $a$, for which $\lim_{x\to 0}\left(\frac{1-\cos(ax)\cos((a+1)x)\cos((a+2)x)}{\sin^2((a+1)x)}\right) = 2$, is:

  1. (1) $-2$
  2. (2) $1$
  3. (3) $-1$
  4. (4) $\frac{5}{4}$
Show answer and solution

Answer: Option (3)

Key idea: Limits using Taylor series expansion and product of roots

  1. Use Taylor expansion: $\cos(\theta) = 1 - \frac{\theta^2}{2} + O(\theta^4)$ and $\sin(\theta) = \theta + O(\theta^3)$
  2. Expand the product: $\cos(ax)\cos((a+1)x)\cos((a+2)x) = 1 - \frac{x^2}{2}(3a^2 + 6a + 5) + O(x^4)$
  3. Numerator becomes: $1 - \cos(ax)\cos((a+1)x)\cos((a+2)x) = \frac{x^2}{2}(3a^2 + 6a + 5) + O(x^4)$
  4. Denominator: $\sin^2((a+1)x) = (a+1)^2x^2 + O(x^4)$
  5. Evaluate limit: $\lim_{x\to 0} \frac{\frac{x^2}{2}(3a^2 + 6a + 5)}{(a+1)^2x^2} = \frac{3a^2 + 6a + 5}{2(a+1)^2} = 2$
  6. Solve: $3a^2 + 6a + 5 = 4(a+1)^2 \Rightarrow a^2 + 2a - 1 = 0$
  7. Product of roots using Vieta's formula: $\frac{-1}{1} = -1$

5 April 2026 Shift 2, Q17 · Q17 · MCQ · Limits and Derivatives · Hard

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x$, $x \in \mathbf{R}$ is :

  1. (1) 0
  2. (2) 2
  3. (3) 3
  4. (4) 1
Show answer and solution

Answer: Option (3)

Key idea: Limits using Taylor series expansion and analyzing transcendental equations

  1. Step 1: Find $f(x)$ by substituting $t = xy$ and using Taylor series: $1 - \cos t = \frac{t^2}{2} + O(t^4)$ and $\tan t = t + O(t^3)$
  2. Step 2: Expand $(1 - \cos t)\tan t = \frac{t^3}{2} + O(t^5)$, so $f(x) = \lim_{t \to 0} \frac{x^3 \cdot \frac{t^3}{2}}{t^3} = \frac{x^3}{2}$
  3. Step 3: Solve $\frac{x^3}{2} = \sin x$ or equivalently $x^3 = 2\sin x$. Define $g(x) = x^3 - 2\sin x$
  4. Step 4: Check $x = 0$: $g(0) = 0$, so $x = 0$ is a solution
  5. Step 5: Analyze $g'(x) = 3x^2 - 2\cos x$. For $x > 0$, $g$ decreases from 0, reaches a negative minimum, then increases to $+\infty$, giving one positive root
  6. Step 6: By odd symmetry $g(-x) = -g(x)$, there is also one negative root
  7. Step 7: Total solutions: $x = 0$, one positive, one negative = **3 solutions**

6 April 2026 Shift 1, Q17 · Q17 · MCQ · Limits and Derivatives · Medium

The value of $\lim_{x \to 0} \left(\frac{x^2 \sin^2 x}{x^2 - \sin^2 x}\right)$ is:

  1. (1) 2
  2. (2) 3
  3. (3) 4
  4. (4) 1
Show answer and solution

Answer: Option (2)

Key idea: Limits using Taylor series expansion and standard limit formulas

  1. Step 1: Identify the indeterminate form $\frac{0}{0}$ when $x \to 0$
  2. Step 2: Use Taylor series: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - ...$
  3. Step 3: Find $\sin^2 x = x^2 - \frac{x^4}{3} + ...$ (higher order terms)
  4. Step 4: Calculate $x^2 - \sin^2 x = x^2 - (x^2 - \frac{x^4}{3} + ...) = \frac{x^4}{3} + ...$
  5. Step 5: Rewrite limit as $\lim_{x \to 0} \frac{\sin^2 x/x^2}{(x^2 - \sin^2 x)/x^4} = \frac{1}{1/3} = 3$

6 April 2026 Shift 2, Q16 · Q16 · MCQ · Limits and Derivatives

Let $\lim_{x \to 2} \frac{(\tan(x-2))(r\sqrt{2}+(p-2)x-2p)}{(x-2)^2} = 5$ for some r, p $\in$ R. If the set of all possible values of q, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

  1. (1) 11
  2. (2) 13
  3. (3) 17
  4. (4) 21
Show answer and solution

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

FAQs

Frequently asked questions

How many questions come from Limits in JEE Main?

In the 9 shifts of JEE Main 2026 Session 2 we found 5 questions on it, about 0.6 per shift.

Where can I practise Limits PYQs for free?

On this page, with solutions. For more, use NTA Abhyas and our free mock tests.

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