PYQ Vault

JEE Mains Maths · Limits and Continuity

Limits via Derivatives and Integrals

Limits that are derivatives in disguise, and limits of integrals with a variable limit, settled by differentiating.

All Limits and Continuity notesJEE Mains Maths strategy

Why this matters

Fifteen PYQs, ten of them multiple choice. Eight recognise a difference quotient or use L'Hospital's rule, often with f and f′ given only at one point; seven have an integral with a variable limit, differentiated by the Leibniz rule. Two ideas cover the page.

Concept 1 of 2: Limits that are derivatives

lim⁡h→0f(a+h)−f(a)h\lim_{h\to0}\frac{f(a+h)-f(a)}h is f′(a)f'(a) by definition. Rearrange a limit into difference quotients and read off derivatives. For any 0/0 form of differentiable functions, L'Hospital's rule replaces the ratio by the ratio of derivatives.

Definition

  • f′(a)=lim⁡h→0f(a+h)−f(a)h=lim⁡x→af(x)−f(a)x−af'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.
  • lim⁡h→0f(a+h)−f(a−h)h=2f′(a)\lim_{h\to0}\frac{f(a+h)-f(a-h)}h=2f'(a).
  • L'Hospital: for 00\frac00 or ∞∞\frac\infty\infty, lim⁡fg=lim⁡f′g′\lim\frac fg=\lim\frac{f'}{g'}.

Derivative as a limit

f′(a)=lim⁡x→af(x)−f(a)x−af'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

Worked example

Find lim⁡x→1x10−1x−1\lim_{x\to1}\frac{x^{10}-1}{x-1}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q60Moderate

Example 1 · Limits and Continuity · Limits via Derivatives and Integrals

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 :

Check the form before L'Hospital

L'Hospital's rule needs 00\frac00 or ∞∞\frac\infty\infty. Applied to a determinate form it gives a wrong answer.
Drill 7 more on limits that are derivatives

Concept 2 of 2: Limits of integrals

An integral with a variable limit, such as ∫0x2g(t) dt\int_0^{x^2}g(t)\,dt, tends to 0 as x→0x\to0, so a ratio with a power of xx is 0/0. Differentiate top and bottom: by the Leibniz rule the top becomes g(x2)⋅2xg(x^2)\cdot2x. Repeat if it is still 0/0.

Definition

  • ddx∫au(x)g(t) dt=g(u(x)) u′(x)\frac d{dx}\int_a^{u(x)}g(t)\,dt=g(u(x))\,u'(x).
  • Use L'Hospital on ∫0u(x)gxn\frac{\int_0^{u(x)}g}{x^n}.
  • Near 0, ∫0xg(t) dt≈g(0)x\int_0^xg(t)\,dt\approx g(0)x.

Leibniz rule

ddx∫au(x)g(t) dt=g(u(x)) u′(x)\frac{d}{dx}\int_a^{u(x)}g(t)\,dt=g\big(u(x)\big)\,u'(x)

Worked example

Find lim⁡x→01x∫0xcos⁡t2 dt\lim_{x\to0}\frac1x\int_0^x\cos t^2\,dt.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q158Moderate

Example 2 · Limits and Continuity · Limits via Derivatives and Integrals

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

Chain rule on the limit

Differentiating ∫0x2g(t) dt\int_0^{x^2}g(t)\,dt gives g(x2)⋅2xg(x^2)\cdot2x, not g(x2)g(x^2). The factor u′(x)u'(x) is easy to drop.
Drill 6 more on limits of integrals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Limits that are derivatives

    Derivative as a limit

    f′(a)=lim⁡x→af(x)−f(a)x−af'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}
  • Limits of integrals

    Leibniz rule

    ddx∫au(x)g(t) dt=g(u(x)) u′(x)\frac{d}{dx}\int_a^{u(x)}g(t)\,dt=g\big(u(x)\big)\,u'(x)

Watch out for (2)

Test yourself on Limits and Continuity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Limits and Continuity chapter testDrill all 15 questions