JEE Maths — Limits Continuity and Differentiability…

Chapter Overview & Weightage

Limits, Continuity, and Differentiability (LCD) is one of the most consistently tested chapters in JEE Main and Advanced. It forms the backbone of Calculus — and Calculus as a whole contributes roughly 35–40% of the Maths paper.

Weightage snapshot: LCD alone accounts for 8–10% of JEE Main Maths (roughly 2–3 questions per paper). In JEE Advanced, it appears in both objective and paragraph-type questions. BITSAT dedicates similar weightage. This is a must-score chapter — medium effort, high return.

Year	JEE Main Questions	Marks	Key Topics Tested
2024	2–3	8–12	L’Hôpital, continuity check, differentiability
2023	2	8	Standard limits, MVT application
2022	3	12	0/0 forms, piecewise functions
2021	2	8	Sandwich theorem, left/right limits
2020	2–3	8–12	L’Hôpital, continuity at a point

The pattern is stable. Expect one pure limit evaluation, one continuity/differentiability check on a piecewise function, and occasionally one MVT-based question in Advanced.

Key Concepts You Must Know

Prioritised by how often they appear in PYQs:

Standard limits — $\lim_{x \to 0} \frac{\sin x}{x} = 1$ , $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ , $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$ are the building blocks of 70% of limit questions
L’Hôpital’s Rule — applies only to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate forms; misapplying it to other forms is an exam trap
1^∞ indeterminate form — evaluated as $e^{\lim f(x) \cdot g(x)}$ where the expression is $[f(x)]^{g(x)} \to 1^\infty$ ; this appears in almost every recent JEE Main paper
Continuity at a point — left-hand limit = right-hand limit = function value; piecewise functions require checking at the joining point
Differentiability implies continuity (but not vice versa) — the converse direction is where most questions live
Left and right derivatives — for $|x|$ , $[x]$ (greatest integer function), and piecewise definitions
Mean Value Theorem (MVT) / Rolle’s Theorem — JEE Advanced uses these for existence proofs and inequality problems
Sandwich (Squeeze) Theorem — appears when direct substitution and L’Hôpital both fail; useful for limits involving $\sin(1/x)$

Important Formulas

\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{\tan x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

When to use: Any limit where you see $\sin$ , $\cos$ , or $\tan$ with the argument $\to 0$ . Rewrite the argument so it matches the denominator, then these kick in directly.

\lim_{x \to 0} \frac{e^x - 1}{x} = 1 \qquad \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 \qquad \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

When to use: Whenever exponentials or logs appear in a $\frac{0}{0}$ form. These are faster than L’Hôpital for these specific structures.

\lim_{x \to a} [f(x)]^{g(x)} \text{ where } f(x) \to 1, g(x) \to \infty

= e^{\displaystyle\lim_{x \to a} [f(x)-1] \cdot g(x)}

When to use: Recognise the $1^\infty$ pattern immediately — it shows up as $\left(1 + \frac{k}{x}\right)^x$ type expressions or disguised versions of them.

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

provided the latter limit exists.

When to use: After confirming the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Can be applied repeatedly. If it cycles, switch to Taylor series.

Continuous at $x = a$ :

\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)

Differentiable at $x = a$ :

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \quad \text{(left and right derivatives must be equal)}

Rolle’s Theorem: If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f(a) = f(b)$ , then $\exists \, c \in (a,b)$ such that $f'(c) = 0$ .

Mean Value Theorem: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then:

f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \in (a,b)

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (January, Shift 1)

Question: $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Solution:

This is $\frac{0}{0}$ at $x = 0$ . We have two clean approaches — L’Hôpital or Taylor expansion. Taylor is faster here.

We know $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

So $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + \cdots}{x^2} = \lim_{x \to 0} \left(\frac{1}{2} + \frac{x}{6} + \cdots\right) = \boxed{\frac{1}{2}}

Taylor expansion around $x = 0$ is almost always faster than repeated L’Hôpital for polynomial-looking limits. Keep the expansions for $e^x$ , $\ln(1+x)$ , $\sin x$ , $\cos x$ memorised to the $x^3$ term.

PYQ 2 — JEE Main 2023 (April, Shift 2)

Question: Find the value of $k$ if $f(x)$ is continuous at $x = 0$ , where:

f(x) = \begin{cases} \dfrac{\sqrt{1+kx} - \sqrt{1-kx}}{x} & x \neq 0 \\ 4 & x = 0 \end{cases}

Solution:

For continuity at $x = 0$ , we need $\lim_{x \to 0} f(x) = f(0) = 4$ .

Rationalise by multiplying numerator and denominator by $\sqrt{1+kx} + \sqrt{1-kx}$ :

\lim_{x \to 0} \frac{(\sqrt{1+kx})^2 - (\sqrt{1-kx})^2}{x\left(\sqrt{1+kx} + \sqrt{1-kx}\right)}

= \lim_{x \to 0} \frac{(1+kx) - (1-kx)}{x\left(\sqrt{1+kx} + \sqrt{1-kx}\right)} = \lim_{x \to 0} \frac{2kx}{x\left(\sqrt{1+kx} + \sqrt{1-kx}\right)}

= \lim_{x \to 0} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1 + 1} = k

Setting $k = 4$ , we get continuity. $\boxed{k = 4}$

A very common error here is forgetting to check whether the limit equals $f(0)$ . Students find the limit correctly but forget the third condition of continuity — the function value must equal the limit.

PYQ 3 — JEE Advanced 2022 (Paper 1)

Question: Let $f(x) = |x - 1| + |x + 1|$ . Check differentiability at $x = 1$ and $x = -1$ .

Solution:

We first write $f(x)$ as a piecewise function by considering the sign of each absolute value piece.

f(x) = \begin{cases} -2x & x < -1 \\ 2 & -1 \leq x \leq 1 \\ 2x & x > 1 \end{cases}

At $x = 1$ :

Left derivative: $\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{2 - 2}{h} = 0$
Right derivative: $\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{2(1+h) - 2}{h} = 2$

Since $0 \neq 2$ , not differentiable at $x = 1$ .

At $x = -1$ :

Left derivative: $\lim_{h \to 0^-} \frac{f(-1+h) - f(-1)}{h} = \lim_{h \to 0^-} \frac{-2(-1+h) - 2}{h} = \frac{2 - 2h - 2}{h} \to -2$
Right derivative: $\lim_{h \to 0^+} \frac{f(-1+h) - f(-1)}{h} = \lim_{h \to 0^+} \frac{2 - 2}{h} = 0$

Since $-2 \neq 0$ , not differentiable at $x = -1$ .

$f$ is continuous everywhere but non-differentiable at $x = \pm 1$ . This is the classic example of “continuous but not differentiable.”

Difficulty Distribution

For JEE Main papers from 2020–2024:

Difficulty	Percentage of LCD Questions	What to Expect
Easy	40%	Standard limit evaluation, continuity of simple piecewise
Medium	45%	$1^\infty$ forms, differentiability of $\\|f(x)\\|$ , parameter-finding
Hard	15%	MVT applications, limits using Taylor + L’Hôpital combined

In JEE Advanced, the difficulty distribution shifts — roughly 60% of LCD questions are medium-hard. Advanced frequently tests MVT in the context of proving inequalities, which feels like a different chapter until you recognise the structure.

Expert Strategy

How toppers approach this chapter

First, master the standard limits cold. The six-to-eight standard limits ( $\sin x/x$ , $e^x-1/x$ , etc.) should be instant recall. When you see a limit, your brain should immediately recognise which standard form to reduce it to.

Build a decision tree for limit evaluation:

Try direct substitution first
If $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → check if factorisation/rationalisation works faster than L’Hôpital
If exponential/log → use standard limits
If $1^\infty$ , $0^0$ , $\infty^0$ → convert to $e^{\text{limit}}$ form
If nothing works → L’Hôpital or Taylor

For continuity/differentiability problems, always draw a rough sketch of the function. For absolute value functions and piecewise definitions, a 10-second sketch tells you exactly where to check and what to expect. Students who skip this step spend 3x longer on the algebra.

For piecewise functions, the protocol is mechanical: identify the joining point, compute left limit, right limit, and function value separately, then compare. Never skip a step here.

MVT/Rolle’s questions in Advanced usually ask you to show a root exists or prove an inequality. The approach: set up an auxiliary function $g(x)$ , verify the conditions of the theorem, then conclude. Practice 5–6 of these and the pattern becomes obvious.

Common Traps

Trap 1: Applying L’Hôpital to non-indeterminate forms. If the form is $\frac{2}{0}$ (not $\frac{0}{0}$ ), the limit is $\pm\infty$ — you cannot use L’Hôpital. Always verify the form before differentiating.

Trap 2: Forgetting that differentiability $\Rightarrow$ continuity, but not the reverse. $f(x) = |x|$ is continuous at $x = 0$ but not differentiable. Questions often give a “continuous function” and ask for differentiability — don’t assume continuity implies differentiability.

Trap 3: The $[x]$ (greatest integer function) and $\{x\}$ (fractional part) ambush. At integer points, $[x]$ has a jump discontinuity, making it non-differentiable there. For $f(x) = x[x]$ , always check $x = 0, 1, 2, \ldots$ separately. Many students apply standard differentiation rules at these points and get the wrong answer.

Trap 4: The $1^\infty$ form disguised. $\lim_{x \to 0} (\cos x)^{1/x^2}$ looks unfamiliar, but it’s $1^\infty$ . Write it as $e^{\frac{1}{x^2} \ln(\cos x)}$ , then expand $\ln(\cos x) \approx -\frac{x^2}{2}$ for small $x$ . Answer: $e^{-1/2}$ . This appeared in JEE Advanced 2019.

Trap 5: Confusing the condition for Rolle’s Theorem. Rolle’s requires $f(a) = f(b)$ . MVT does not. Students mix these up under time pressure and either apply the wrong theorem or miss that the condition isn’t satisfied.