How much of JEE Main comes from Limits Continuity and Differentiability?

JEEGenie estimates about 6% of JEE Main marks come from Limits Continuity and Differentiability, based on recent papers. The share varies from year to year.

Which formulas matter most in Limits Continuity and Differentiability?

The Limits Continuity and Differentiability sheet on this page has 24 formula entries, starting with Limit of a Function, Continuity at a Point, Differentiability at a Point, Standard Limits: Trigonometric, Standard Limits: Exponential and Logarithmic. Each one comes with when to use it and the trap to avoid.

What topics does Limits Continuity and Differentiability cover for JEE?

Limits Continuity and Differentiability covers Limits, Continuity, Differentiability, Derivatives, Rolle's and Mean Value Theorem, L'Hopital's Rule.

How should I revise Limits Continuity and Differentiability before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Limits Continuity and Differentiability questions on JEEGenie.

How many previous-year questions are there on Limits Continuity and Differentiability?

JEEGenie has 1008 previous-year questions on Limits Continuity and Differentiability, and this page shows a few with worked explanations.

Is practice on Limits Continuity and Differentiability free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Limits Continuity and Differentiability for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 6% of JEE Main marks come from Limits Continuity and Differentiability. This page covers 24 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Assuming $f (a)$ must equal the limit; the limit exists even if $f (a)$ is undefined.

Key points to know

Limit of a Function. The limit of $f (x)$ as $x$ approaches $a$ is $L$ , written $\lim_{x \to a} f (x) = L$ , if $f (x)$ gets arbitrarily close to $L$ for all $x$ sufficiently close to $a$ (but not equal to $a$ ).
Continuity at a Point. A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f (x) = f (a)$ . This requires $f (a)$ defined, the limit exists, and they are equal.
Differentiability at a Point. A function $f$ is differentiable at $x = a$ if the derivative $f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ exists (finite). This implies continuity at $a$ .
Types of Discontinuity. Discontinuities are classified as removable (limit exists but not equal to function value), jump (one-sided limits exist but differ), or infinite (limit is $\pm \infty$ ).
Continuity on an Interval. A function is continuous on an interval if it is continuous at every point in the interval. For closed intervals, continuity at endpoints means one-sided continuity.
Differentiability and Continuity Relationship. If a function is differentiable at a point, it is necessarily continuous there. The converse is false; continuity does not imply differentiability.
Mean Value Theorem (MVT). If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists $c \in (a, b)$ such that $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ .
Continuity of Composite Functions. If $g$ is continuous at $a$ and $f$ is continuous at $g (a)$ , then the composite $f \circ g$ is continuous at $a$ .

Formula sheet

Name	Formula	When to use
Limit of a Function	$\lim_{x \to a} f (x) = L$	When analyzing the behavior of a function near a point, especially for continuity and differentiability checks.
Continuity at a Point	$\lim_{x \to a} f (x) = f (a)$	When determining if a function has no jumps, holes, or asymptotes at a point, often for piecewise functions.
Differentiability at a Point	$f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$	When checking if a function has a well-defined tangent or rate of change at a point, especially for piecewise functions.
Standard Limits: Trigonometric	$\lim_{x \to 0} \frac{\sin x}{x} = 1$ , $\lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}$	When evaluating limits involving sine, cosine, or tangent as the variable approaches 0, often after algebraic manipulation.
Standard Limits: Exponential and Logarithmic	$\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$ , $\lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1$ , $\lim_{x \to \infty} (1 + \frac{1}{x})^{x} = e$	When evaluating limits involving exponential or logarithmic functions, especially in indeterminate forms like $0 / 0$ or $1^{\infty}$ .
L'Hôpital's Rule	$\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ for $\frac{0}{0}$ or $\frac{\infty}{\infty}$	When direct substitution yields $0 / 0$ or $\infty / \infty$ , especially with trigonometric, exponential, or logarithmic functions.
Rationalization for Limits	$\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a} = \frac{1}{2 \sqrt{a}}$ after rationalization	When evaluating limits with square roots, cube roots, or other radicals that give $0 / 0$ form.
Squeeze Theorem (Sandwich Theorem)	$g (x) \leq f (x) \leq h (x)$ and $\lim g = \lim h = L ⟹ \lim f = L$	When direct evaluation is impossible and the function is bounded between two simpler functions with known limits, e.g., $\lim_{x \to 0} x^{2} \sin (1 / x)$ .
Types of Discontinuity	Removable: $\lim_{x \to a} f (x) = L$ exists (finite) but $f (a)$ is undefined or $f (a) \neq L$ ; Jump: $\lim_{x \to a^{-}} f (x) \neq \lim_{x \to a^{+}} f (x)$	When analyzing the nature of a break in a function, especially for piecewise or rational functions.
Derivative as a Limit	$f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$	When deriving derivative formulas from first principles or checking differentiability at a point.
Chain Rule for Derivatives	$\frac{d}{d x} f (g (x)) = f^{'} (g (x)) \cdot g^{'} (x)$	When differentiating composite functions, such as $\sin (\ln x)$ , $e^{x^{2}}$ , or $\sqrt{1 + x^{2}}$ .
L'Hôpital's Rule for $1^{\infty}$ , $0^{0}$ , $\infty^{0}$ Forms	$\lim f (x)^{g (x)} = e^{\lim g (x) \ln f (x)}$	When evaluating limits of the form $f (x)^{g (x)}$ where $f (x) \to 1$ and $g (x) \to \infty$ , or similar.
Continuity on an Interval	Continuous on $[a, b]$ : continuous at each $x \in (a, b)$ , $\lim_{x \to a^{+}} f (x) = f (a)$ , $\lim_{x \to b^{-}} f (x) = f (b)$	When applying theorems like Intermediate Value Theorem or Extreme Value Theorem that require interval continuity.
Intermediate Value Theorem (IVT)	$f$ continuous on $[a, b]$ , $k$ between $f (a)$ and $f (b)$ $⟹ \exists c \in [a, b] : f (c) = k$ (and $c \in (a, b)$ if $k$ is strictly between)	When proving existence of roots or solutions to equations, especially for continuous functions.
Differentiation from First Principles	$f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$	When deriving derivative formulas for new functions or checking differentiability at a specific point.
Differentiability and Continuity Relationship	$f$ differentiable at $a ⟹ f$ continuous at $a$	When analyzing the smoothness of a function; differentiability ensures no corners or cusps.
Standard Derivatives	$\frac{d}{d x} x^{n} = n x^{n - 1}$ , $\frac{d}{d x} \sin x = \cos x$ , $\frac{d}{d x} e^{x} = e^{x}$ , $\frac{d}{d x} \ln x = \frac{1}{x}$	When differentiating basic functions, especially in combination with product, quotient, and chain rules.
Implicit Differentiation	$\frac{d}{d x} f (y) = f^{'} (y) \cdot \frac{d y}{d x}$	When differentiating equations where $y$ cannot be easily expressed as a function of $x$ , e.g., $x^{2} + y^{2} = 1$ .
Logarithmic Differentiation	$\ln y = g (x) \ln f (x) ⟹ \frac{1}{y} \frac{d y}{d x} = g^{'} (x) \ln f (x) + g (x) \frac{f^{'} (x)}{f (x)}$	When differentiating functions like $x^{x}$ , $\frac{(x + 1)^{3} \sqrt{x}}{(x - 1)^{2}}$ , or $\sin x^{\cos x}$ .
Mean Value Theorem (MVT)	$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ for some $c \in (a, b)$	When proving properties of functions, such as monotonicity or existence of points with specific derivative values.
Rolle's Theorem	$f (a) = f (b) ⟹ \exists c \in (a, b) : f^{'} (c) = 0$	When proving that a function has a stationary point between two equal values, often used in root analysis.
Evaluating Limits Using Series Expansion	$\sin x = x - \frac{x^{3}}{6} + O (x^{5})$ , $e^{x} = 1 + x + \frac{x^{2}}{2} + O (x^{3})$	When L'Hôpital's rule leads to repeated differentiation or when the limit involves trigonometric/exponential functions near 0.
Handling Absolute Value in Limits and Derivatives	$∣ x ∣ = {\begin{cases} x & x \geq 0 \\ - x & x < 0 \end{cases}$	When evaluating limits or derivatives of functions like $∣ x - 1 ∣$ , $∣ \sin x ∣$ , or $∣ x ∣^{3}$ .
Continuity of Composite Functions	$\lim_{x \to a} f (g (x)) = f (\lim_{x \to a} g (x)) = f (g (a))$	When analyzing continuity of functions like $\sin (1 / x)$ or $e^{\sqrt{x}}$ , especially at boundary points.

Open the Limits Continuity and Differentiability formula page

What JEE asks

Limit of a Function. JEE often tests existence of limits using piecewise functions or trigonometric expressions, requiring evaluation of left and right limits.
Continuity at a Point. JEE frequently asks to find constants for piecewise functions to ensure continuity, using limit equality.
Differentiability at a Point. JEE tests differentiability by evaluating left and right derivatives for piecewise functions, often with absolute values or mod functions.
Standard Limits: Trigonometric. JEE often combines these with L'Hôpital's rule or series expansions, testing manipulation like $\frac{\sin (a x)}{b x}$ .
Standard Limits: Exponential and Logarithmic. JEE tests these in combination with L'Hôpital's rule or by converting to standard forms, often in limits with $e$ and $\ln$ .
L'Hôpital's Rule. JEE often requires multiple applications of L'Hôpital's rule, or combining it with algebraic simplification to avoid repeated differentiation.
Rationalization for Limits. JEE tests this in limits with expressions like $\frac{\sqrt{1 + x} - 1}{x}$ or $\frac{\sqrt{x^{2} + 1} - x}{x}$ .
Squeeze Theorem (Sandwich Theorem). JEE often uses this for limits involving oscillatory functions like $\sin (1 / x)$ or $\cos (1 / x)$ multiplied by a term that goes to 0.
Types of Discontinuity. JEE asks to identify the type of discontinuity from graphs or function definitions, often with rational functions.
Derivative as a Limit. JEE sometimes asks to find derivatives using the limit definition, especially for functions like $∣ x ∣$ or $\sin x$ .
Chain Rule for Derivatives. JEE tests chain rule in combination with other rules, often in implicit differentiation or parametric equations.
L'Hôpital's Rule for $1^{\infty}$ , $0^{0}$ , $\infty^{0}$ Forms. JEE often tests this with limits like $\lim_{x \to 0} (1 + \sin x)^{1 / x}$ or $\lim_{x \to 0} x^{x}$ .
Continuity on an Interval. JEE asks to find intervals of continuity for piecewise or rational functions, often involving domain restrictions.
Intermediate Value Theorem (IVT). JEE tests IVT in theoretical problems, often asking to show a root exists in an interval without finding it.
Differentiation from First Principles. JEE sometimes asks to find derivatives of functions like $\sqrt{x}$ or $\frac{1}{x}$ using first principles.
Differentiability and Continuity Relationship. JEE often tests this by asking to find points where a function is continuous but not differentiable, like $∣ x ∣$ or $x^{2 / 3}$ .
Standard Derivatives. JEE tests these in all derivative problems, often requiring quick recall for composite functions.
Implicit Differentiation. JEE tests implicit differentiation in problems involving circles, ellipses, or relations like $\sin (x y) = x$ .
Logarithmic Differentiation. JEE often uses this for functions with variable exponents or multiple factors, simplifying the differentiation process.
Mean Value Theorem (MVT). JEE tests MVT in theoretical problems, often asking to find $c$ or prove inequalities.
Rolle's Theorem. JEE uses Rolle's theorem to show that $f^{'} (x) = 0$ has a root between two consecutive roots of $f$ , or to bound the number of real roots of a polynomial.
Evaluating Limits Using Series Expansion. JEE often tests this for limits like $\lim_{x \to 0} \frac{\sin x - x}{x^{3}}$ or $\lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}}$ .
Handling Absolute Value in Limits and Derivatives. JEE frequently tests differentiability of $∣ x ∣$ at 0, or limits involving $\frac{∣ x ∣}{x}$ .
Continuity of Composite Functions. JEE tests this in problems involving piecewise functions where one function is applied to another.

Common traps

Assuming $f (a)$ must equal the limit; the limit exists even if $f (a)$ is undefined. Limit of a Function
Forgetting to check that $f (a)$ is defined; a hole in the graph means discontinuity. Continuity at a Point
Assuming differentiability from continuity; continuity does not guarantee differentiability (e.g., $∣ x ∣$ at $x = 0$ ). Differentiability at a Point
Applying $\lim_{x \to 0} \frac{\sin x}{x} = 1$ when the argument is not $x$ (e.g., $\frac{\sin 2 x}{x}$ requires adjustment). Standard Limits: Trigonometric
Applying $\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$ when the exponent is not $x$ (e.g., $\frac{e^{2 x} - 1}{x}$ needs factoring). Standard Limits: Exponential and Logarithmic
Applying L'Hôpital's rule when the limit is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ; it gives incorrect results otherwise. L'Hôpital's Rule
Forgetting to multiply both numerator and denominator by the conjugate, leading to an incorrect expression. Rationalization for Limits
Using the theorem without verifying the inequality holds for all $x$ near $a$ (except possibly at $a$ ). Squeeze Theorem (Sandwich Theorem)
Calling a discontinuity 'removable' when the limit does not exist; it must exist for removal. Types of Discontinuity
Using $h \to 0$ but forgetting that $h$ can be negative; the limit must be two-sided. Derivative as a Limit
Forgetting to multiply by the derivative of the inner function, e.g., differentiating $\sin (x^{2})$ as $\cos (x^{2})$ instead of $2 x \cos (x^{2})$ . Chain Rule for Derivatives
Applying L'Hôpital directly to $1^{\infty}$ without taking logs; it's not a $0 / 0$ or $\infty / \infty$ form. L'Hôpital's Rule for $1^{\infty}$ , $0^{0}$ , $\infty^{0}$ Forms
Assuming continuity on an open interval implies continuity on the closed interval; endpoints need separate checks. Continuity on an Interval
Applying IVT when $f$ is not continuous on the closed interval; the theorem fails. Intermediate Value Theorem (IVT)

Topics in Limits Continuity and Differentiability

Limits
Continuity
Differentiability
Derivatives
Rolle's and Mean Value Theorem
L'Hopital's Rule

Previous-year questions on Limits Continuity and Differentiability

JEEGenie has 1008 previous-year questions on Limits Continuity and Differentiability. A few worked examples follow.

Year	Questions
2026	34
2025	49
2024	74
2023	53
2022	74
2021	92
2020	66
2019	114
2018	30
2017	26
2016	25
2015	14
2014	22
2013	19
2012	21
2011	8
2010	10
2009	6
2008	11
2007	7
2006	13
2005	14
2004	13
2003	13
2002	17
2001	6
2000	6
1999	8
1998	8
1997	14
1996	8
1995	7
1994	5
1993	9
1992	3
1991	7
1990	7
1989	7
1988	5
1987	8
1986	9
1985	11
1984	6
1983	19
1982	10
1981	11
1980	4
1979	3
1978	2

Question 1 hard

\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m

, then

a + b + m

is equal to :

Answer: B

Explanation. For the limit to be finite and nonzero, the numerator must also tend to

0

x \to 2

. Since

\sqrt{x - 1} - 1 \to 0

and

\log_{e} (x - 1) \to 0

, the denominator behaves like

(\sqrt{x - 1} - 1) \log_{e} (x - 1) \sim \frac{x - 2}{2} \cdot (x - 2) = \frac{(x - 2)^{2}}{2} .

Thus we need

\sin (x^{3} - 5 x^{2} + a x + b) \sim k (x - 2)^{2}

for some constant

k

. For this, the argument of sine must vanish at

x = 2

8 - 20 + 2 a + b = 0 ⟹ 2 a + b = 12.

Also, for the leading term to be quadratic, the derivative of the argument at

x = 2

must be zero:

3 x^{2} - 10 x + a ∣_{x = 2} = 12 - 20 + a = a - 8 = 0 ⟹ a = 8.

Then

2 (8) + b = 12 ⟹ b = - 4

. Now expand the argument near

x = 2

: Let

t = x - 2

. Then

x^{3} - 5 x^{2} + 8 x - 4 = (2 + t)^{3} - 5 (2 + t)^{2} + 8 (2 + t) - 4.

Compute:

(8 + 12 t + 6 t^{2} + t^{3}) - 5 (4 + 4 t + t^{2}) + 16 + 8 t - 4

= 8 + 12 t + 6 t^{2} + t^{3} - 20 - 20 t - 5 t^{2} + 16 + 8 t - 4

= (8 - 20 + 16 - 4) + (12 - 20 + 8) t + (6 - 5) t^{2} + t^{3} = 0 + 0 t + t^{2} + t^{3} .

So the argument is

t^{2} + t^{3} = (x - 2)^{2} + (x - 2)^{3}

. Thus

\sin ((x - 2)^{2} + (x - 2)^{3}) \sim (x - 2)^{2} .

Denominator:

\sqrt{x - 1} - 1 = \frac{x - 2}{\sqrt{x - 1} + 1} \sim \frac{x - 2}{2},

and

\log_{e} (x - 1) = \log_{e} (1 + (x - 2)) \sim x - 2.

So denominator

\sim \frac{(x - 2)^{2}}{2}

. Therefore

m = \lim_{x \to 2} \frac{(x - 2)^{2}}{\frac{(x - 2)^{2}}{2}} = 2.

Hence

a + b + m = 8 + (- 4) + 2 = 6.

6

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 medium

If the function

f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1

, where

a > 0

, attains its local maximum and local minimum values at p and q , respectively, such that

p^{2} = q

, then

f (3)

is equal to :

Answer: B

Explanation.

f^{'} (x) = 6 x^{2} - 18 a x + 12 a^{2} = 6 (x^{2} - 3 a x + 2 a^{2}) = 6 (x - a) (x - 2 a)

. Critical points:

x = a

and

x = 2 a

. Since

a > 0

a < 2 a

. For cubic with positive leading coefficient, local max at

x = a

, local min at

x = 2 a

. So

p = a

q = 2 a

. Given

p^{2} = q ⟹ a^{2} = 2 a ⟹ a (a - 2) = 0

, so

a = 2

(since

a > 0

). Thus

f (x) = 2 x^{3} - 18 x^{2} + 48 x + 1

f (3) = 2 (27) - 18 (9) + 48 (3) + 1 = 54 - 162 + 144 + 1 = 37

. The answer is option B.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 easy

The function

f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}

decreases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$
increases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$
decreases in $(- 2, 8)$ and increases in $(- \infty, - 2) \cup (8, \infty)$
decreases in $(- \infty, - 2)$ and increases in $(8, \infty)$

Answer: A

Explanation. We are given

f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8} .

First factor the denominator:

x^{2} - 6 x - 16 = (x - 8) (x + 2) .

f (x) = \frac{x}{(x - 8) (x + 2)} .

Differentiate using the quotient rule. Let

N = x, D = x^{2} - 6 x - 16.

Then

f^{'} (x) = \frac{N^{'} D - N D^{'}}{D^{2}} = \frac{1 \cdot (x^{2} - 6 x - 16) - x (2 x - 6)}{(x^{2} - 6 x - 16)^{2}} .

Simplify the numerator:

x^{2} - 6 x - 16 - 2 x^{2} + 6 x = - x^{2} - 16.

Thus

f^{'} (x) = \frac{- x^{2} - 16}{(x^{2} - 6 x - 16)^{2}} = - \frac{x^{2} + 16}{(x^{2} - 6 x - 16)^{2}} .

For all

x

in the domain, the denominator is positive because it is a square, and

x^{2} + 16 > 0

. Hence

f^{'} (x) < 0

for every

x \in R - {- 2, 8}

. Therefore

f

is strictly decreasing on each interval of its domain:

(- \infty, - 2), (- 2, 8), (8, \infty) .

f

decreases in

(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty) .

The tempting wrong option is (C), which says the function decreases in

(- 2, 8)

but increases in

(- \infty, - 2) \cup (8, \infty)

. That would require

f^{'} (x) > 0

on those outer intervals, but

f^{'} (x)

is always negative wherever it is defined. Hence (C) is incorrect. Therefore the correct option is (A). (A) decreases in

(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)

From NTA/IIT papers: JEE Main 2024 · 29 Jan Shift 2

Question 4 medium

Let

a_{1}, a_{2}, a_{3}, \dots, a_{n}

n

positive consecutive terms of an arithmetic progression. If

d > 0

is its common difference, then

\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots \dots \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}})

$\frac{1}{\sqrt{d}}$
1
0
$\sqrt{d}$

Answer: B

Explanation. **Principle:** Rationalize each term using

(\sqrt{a_{k}} + \sqrt{a_{k + 1}})

and the identity

a_{k + 1} - a_{k} = d

. For consecutive AP terms,

a_{k + 1} = a_{k} + d

, so

\frac{1}{\sqrt{a_{k}} + \sqrt{a_{k + 1}}} = \frac{\sqrt{a_{k + 1}} - \sqrt{a_{k}}}{a_{k + 1} - a_{k}} = \frac{\sqrt{a_{k + 1}} - \sqrt{a_{k}}}{d} .

Summing from

k = 1

n - 1

, the series telescopes:

S = \frac{1}{d} (\sqrt{a_{n}} - \sqrt{a_{1}}) .

Now

a_{n} = a_{1} + (n - 1) d

, so

\sqrt{\frac{d}{n}} S = \frac{1}{\sqrt{d n}} (\sqrt{a_{1} + (n - 1) d} - \sqrt{a_{1}}) .

Divide numerator and denominator by

\sqrt{n}

= \frac{1}{\sqrt{d}} (\sqrt{\frac{a_{1}}{n} + d - \frac{d}{n}} - \sqrt{\frac{a_{1}}{n}}) .

n \to \infty

\frac{a_{1}}{n} \to 0

and

\frac{d}{n} \to 0

, giving

\frac{1}{\sqrt{d}} (\sqrt{d} - 0) = 1.

The other options fail:

\frac{1}{\sqrt{d}}

and

\sqrt{d}

depend on

d

, while the limit is the constant

1

;

0

would require the telescoped difference to vanish, which it does not. Answer: (B)

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

Let

α

be a positive real number. Let

f : R \to R

and

g : (α, \infty) \to R

be the functions defined by

f (x) = \sin (\frac{π x}{12})

and

g (x) = \frac{2 \log_{e} (\sqrt{x} - \sqrt{α})}{\log_{e} (e^{\sqrt{x}} - e^{\sqrt{α}})}

. Then the value of

\lim_{x \to α^{+}} f (g (x))

is ______.

Answer: 0.50

Explanation.

\lim_{x \to α^{+}} g (x) = \lim_{x \to α^{+}} \frac{2 \ln (\sqrt{x} - \sqrt{α})}{\ln (e^{\sqrt{x}} - e^{\sqrt{α}})}

Let

t = \sqrt{x} - \sqrt{α}

. As

x \to α^{+}

t \to 0^{+}

. Also,

\sqrt{x} = \sqrt{α} + t, e^{\sqrt{x}} - e^{\sqrt{α}} = e^{\sqrt{α}} (e^{t} - 1) .

g (x) = \frac{2 \ln t}{\ln (e^{\sqrt{α}} (e^{t} - 1))} = \frac{2 \ln t}{\sqrt{α} + \ln (e^{t} - 1)} .

Now

e^{t} - 1 \sim t

t \to 0

, so

\ln (e^{t} - 1) \sim \ln t

. Hence

g (x) \sim \frac{2 \ln t}{\ln t} = 2.

Thus

\lim_{x \to α^{+}} g (x) = 2.

Therefore,

\lim_{x \to α^{+}} f (g (x)) = f (2) = \sin (\frac{2 π}{12}) = \sin (\frac{π}{6}) = \frac{1}{2} .

0.50

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Limits Continuity and Differentiability?: JEEGenie estimates about 6% of JEE Main marks come from Limits Continuity and Differentiability, based on recent papers. The share varies from year to year.
Which formulas matter most in Limits Continuity and Differentiability?: The Limits Continuity and Differentiability sheet on this page has 24 formula entries, starting with Limit of a Function, Continuity at a Point, Differentiability at a Point, Standard Limits: Trigonometric, Standard Limits: Exponential and Logarithmic. Each one comes with when to use it and the trap to avoid.
What topics does Limits Continuity and Differentiability cover for JEE?: Limits Continuity and Differentiability covers Limits, Continuity, Differentiability, Derivatives, Rolle's and Mean Value Theorem, L'Hopital's Rule.
How should I revise Limits Continuity and Differentiability before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Limits Continuity and Differentiability questions on JEEGenie.
How many previous-year questions are there on Limits Continuity and Differentiability?: JEEGenie has 1008 previous-year questions on Limits Continuity and Differentiability, and this page shows a few with worked explanations.
Is practice on Limits Continuity and Differentiability free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.