GATE DA

Limits, continuity and differentiability

This topic covers how a function of one real variable behaves near a point: whether a limit exists, whether the function is continuous there, and whether it has a derivative. GATE questions test standard limits, L'Hopital's rule, parameters that make a piecewise function continuous or differentiable, and the mean value theorems, and all of them rest on the definitions and the one-way link that differentiability implies continuity.

6 previous-year questions · 6 solved · 5 practice questions

Limits and the standard results

Convention: all functions here are real-valued functions of one real variable, $ln$ is the natural logarithm and angles are in radians. The limit $lim_{x \to a} f (x) = L$ (with $L$ finite) means $f (x)$ gets arbitrarily close to $L$ as $x$ gets close to $a$ (from both sides), whatever $f (a)$ itself is, or even if $f (a)$ is undefined. At an endpoint of the domain only the one-sided limit is used.

LIMIT EXISTS IFF BOTH ONE-SIDED LIMITS AGREE

x \to a lim f (x) = L ⟺ x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = L

Use one-sided limits for $∣ x ∣$ , floor, piecewise functions and any function that changes formula at $a$ .

LIMIT LAWS

x \to a lim [f \pm g] = L \pm M, x \to a lim [f g] = L M, x \to a lim \frac{f}{g} = \frac{L}{M} (M \neq = 0)

Valid when $lim f = L$ and $lim g = M$ both exist and are finite.

STANDARD TRIGONOMETRIC LIMITS

x \to 0 lim \frac{sin x}{x} = 1, x \to 0 lim \frac{tan x}{x} = 1, x \to 0 lim \frac{1 - cos x}{x ^{2}} = \frac{1}{2}

Only for $x$ in radians and $x \to 0$ . For $x \to \infty$ , $sin x / x \to 0$ instead.

STANDARD EXPONENTIAL AND LOGARITHMIC LIMITS

x \to 0 lim \frac{e ^{x} - 1}{x} = 1, x \to 0 lim \frac{a ^{x} - 1}{x} = ln a (a > 0), x \to 0 lim \frac{ln ( 1 + x )}{x} = 1

Each is a $0/0$ form that equals a derivative at $0$ (of $e^{x}$ , $a^{x}$ and $ln (1 + x)$ respectively).

LIMITS DEFINING E AND THE BINOMIAL LIMIT

x \to \pm \infty lim (1 + \frac{1}{x})^{x} = e, x \to 0 lim (1 + x)^{1/ x} = e, x \to 0 lim \frac{( 1 + x ) ^{n} - 1}{x} = n

Here $n$ is any real number. More generally $lim_{x \to \infty} (1 + k / x)^{x} = e^{k}$ for any real $k$ .

THE $1^\INFTY$ FORM

x \to a lim f (x)^{g (x)} = exp (x \to a lim g (x) [f (x) - 1])

Applies when $f (x) > 0$ near $a$ , $f (x) \to 1$ , $g (x) \to \pm \infty$ , and the limit inside the exponential exists.

EXPANSIONS NEAR $X=0$: EXPONENTIAL AND LOGARITHM

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots, ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

The $e^{x}$ series holds for all real $x$ ; the $ln (1 + x)$ series holds for $- 1 < x \leq 1$ .

EXPANSIONS NEAR $X=0$: TRIGONOMETRIC

sin x = x - \frac{x ^{3}}{3 !} + \dots, cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots, tan x = x + \frac{x ^{3}}{3} + \dots

The $sin x$ and $cos x$ series hold for all real $x$ ; the $tan x$ series for $∣ x ∣ < π /2$ . In a $0/0$ limit at $0$ , replace each function by enough terms and compare the lowest powers: for example $\frac{x - sin x}{x ^{3}} = \frac{x ^{3} /6 - \dots}{x ^{3}} \to \frac{1}{6}$ .

Other tools: factorise and cancel for $0/0$ with polynomials, multiply by the conjugate for square-root differences, divide by the highest power for $x \to \infty$ , and use the squeeze theorem: if $g \leq f \leq h$ near $a$ and $g, h \to L$ then $f \to L$ . For example $x sin (1/ x) \to 0$ as $x \to 0$ because $∣ x sin (1/ x) ∣ \leq ∣ x ∣$ .

Indeterminate forms and L'Hopital's rule

The forms $0/0$ , $\infty/\infty$ , $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ , $0^{0}$ and $\infty^{0}$ do not have a value decided by the pieces alone. Reduce each to $0/0$ or $\infty/\infty$ and then apply L'Hopital's rule, or use a standard limit or a series expansion.

L'HOPITAL'S RULE

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

Needs: $f / g$ of the form $0/0$ or $\pm \infty/ \pm \infty$ as $x \to a$ ; $f, g$ differentiable with $g^{'} (x) \neq = 0$ on a punctured neighbourhood of $a$ ; and the limit of $f^{'} / g^{'}$ (the limit on the right of the equation) to exist, finite or infinite. It also holds for one-sided limits and for $a = \pm \infty$ . Differentiate numerator and denominator separately; do not use the quotient rule.

Reducing other forms

$0 \cdot \infty$ : write $f g$ as $f / (1/ g)$ or $g / (1/ f)$ to get $0/0$ or $\infty/\infty$ .
$\infty - \infty$ : combine into one fraction, or multiply and divide by the conjugate.
$1^{\infty}$ , $0^{0}$ , $\infty^{0}$ (base positive near $a$ ): take $ln$ of the expression, find the limit $ℓ$ of the logarithm by L'Hopital's rule, then the answer is $e^{ℓ}$ .
Repeat L'Hopital's rule only while the form remains $0/0$ or $\infty/\infty$ .

GROWTH ORDER

x \to \infty lim \frac{( ln x ) ^{m}}{x ^{p}} = 0, x \to \infty lim \frac{x ^{p}}{e ^{x}} = 0 (p > 0, m > 0)

Exponentials beat powers and powers beat logarithms as $x \to \infty$ .

WORKED EXAMPLE

Find $lim_{x \to \infty} (1 + \frac{3}{x})^{2 x}$ .

Solution

The form is $1^{\infty}$ .
Write it as $[(1 + \frac{3}{x})^{x}]^{2}$ .
Since $(1 + \frac{3}{x})^{x} \to e^{3}$ , the whole expression tends to $(e^{3})^{2} = e^{6}$ .
Check with the $1^{\infty}$ rule: $g (f - 1) = 2 x \cdot \frac{3}{x} = 6$ , so the limit is $e^{6}$ .

Answer: $e^{6}$

EXAM TIP

Before using L'Hopital's rule, confirm the form really is $0/0$ or $\infty/\infty$ . Applying it to a determinate form, such as $lim_{x \to 0} (cos x) / x$ , gives a wrong answer. Also, if the limit of $f^{'} / g^{'}$ does not exist, the rule says nothing about $f / g$ .

Continuity and types of discontinuity

CONTINUITY AT A POINT

f is continuous at a ⟺ x \to a lim f (x) = f (a)

Three things must hold: $f (a)$ is defined, the limit exists and is finite, and the two are equal. Equivalently, left limit = right limit = $f (a)$ .

A function is continuous on an interval if it is continuous at every point of it (at an endpoint, use the one-sided limit). Sums, differences, products and quotients (where the denominator is nonzero) of functions continuous at $a$ are continuous at $a$ ; a composition $f (g (x))$ is continuous at $a$ when $g$ is continuous at $a$ and $f$ is continuous at $g (a)$ . Polynomials, $e^{x}$ , $sin x$ , $cos x$ , $∣ x ∣$ are continuous everywhere, $ln x$ on $x > 0$ , $tan x$ wherever $cos x \neq = 0$ .

In the table below, following common usage in GATE questions, a point where $f$ is undefined but defined on both sides near it is also classified as a discontinuity.

Type	What happens at $a$	Example at $x = 0$
Removable	$lim_{x \to a} f (x)$ exists and is finite but differs from $f (a)$ , or $f (a)$ is undefined	$\frac{sin x}{x}$ (undefined at $0$ , limit $1$ )
Jump	Both one-sided limits exist and are finite but are unequal	$\frac{∣ x ∣}{x}$ (limits $- 1$ and $1$ )
Infinite	At least one one-sided limit is $\pm \infty$	$\frac{1}{x}$
Oscillating	At least one one-sided limit does not exist and is not infinite	$sin \frac{1}{x}$

INTERMEDIATE VALUE THEOREM

f continuous on [a, b], f (a) \neq = f (b), y strictly between f (a) and f (b) \Rightarrow \exists c \in (a, b) : f (c) = y

Used to show a root exists, for example $f (a) f (b) < 0$ gives a root of $f$ in $(a, b)$ .

EXTREME VALUE THEOREM

f continuous on the closed bounded interval [a, b] \Rightarrow f attains a global maximum and a global minimum on [a, b]

Both conditions matter: on the open interval $(0, 1)$ , $f (x) = 1/ x$ is continuous but has no maximum.

EXAM TIP

To make a piecewise function continuous at the join point, set left limit = right limit = value there. That gives one equation per join. Do not forget to evaluate the one-sided limits using the correct formula on each side.

Differentiability and first principles

DERIVATIVE FROM FIRST PRINCIPLES

f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h} = x \to a lim \frac{f ( x ) - f ( a )}{x - a}

$f$ is differentiable at $a$ when this limit exists and is finite.

LEFT AND RIGHT DERIVATIVES

f_{-}^{'} (a) = h \to 0^{-} lim \frac{f ( a + h ) - f ( a )}{h}, f_{+}^{'} (a) = h \to 0^{+} lim \frac{f ( a + h ) - f ( a )}{h}

$f$ is differentiable at $a$ if and only if both exist, are finite and $f_{-}^{'} (a) = f_{+}^{'} (a)$ .

DIFFERENTIABILITY IMPLIES CONTINUITY

f differentiable at a \Rightarrow f continuous at a

The converse is false. Hence a function that is not continuous at $a$ cannot be differentiable at $a$ .

The standard counterexample is $f (x) = ∣ x ∣$ : it is continuous at $0$ , but $f_{-}^{'} (0) = - 1$ and $f_{+}^{'} (0) = 1$ , so it is not differentiable there. Common ways a function fails to be differentiable at a point include a corner (unequal one-sided slopes, e.g. $∣ x ∣$ at $0$ ), a vertical tangent (slope $\pm \infty$ , e.g. $x^{1/3}$ at $0$ ), a cusp (one-sided slopes $- \infty$ and $+ \infty$ , e.g. $x^{2/3}$ at $0$ ), oscillation (e.g. $x sin (1/ x)$ with $f (0) = 0$ , continuous at $0$ but $f (h) / h = sin (1/ h)$ has no limit), and a discontinuity.

Function	At $x = 0$	Reason
$∣ x ∣$	continuous, not differentiable	corner: slopes $- 1$ and $+ 1$
$x^{1/3}$	continuous, not differentiable	vertical tangent, slope $\to \infty$
$x ∣ x ∣$	differentiable, $f^{'} (0) = 0$	$f (h) / h = ∣ h ∣ \to 0$
$x^{2} sin (1/ x)$ , $f (0) = 0$	differentiable, $f^{'} (0) = 0$ , but $f^{'}$ is not continuous at $0$	$f^{'} (x) = 2 x sin (1/ x) - cos (1/ x)$ has no limit as $x \to 0$

WORKED EXAMPLE

Find constants $a$ and $b$ so that $f (x) = a x^{2} + b$ for $x \leq 1$ and $f (x) = 4 x - 1$ for $x > 1$ is differentiable at $x = 1$ .

Solution

Differentiability needs continuity first. Left value: $a + b$ . Right limit: $4 (1) - 1 = 3$ . So $a + b = 3$ .
Left derivative at $1$ : $2 a x$ at $x = 1$ is $2 a$ . Right derivative (using $f (1) = a + b = 3$ ): $lim_{h \to 0^{+}} \frac{4 ( 1 + h ) - 1 - 3}{h} = 4$ . So $2 a = 4$ and $a = 2$ .
Then $b = 3 - a = 1$ .

Answer: $a = 2, b = 1$

Rules and standard derivatives

PRODUCT AND QUOTIENT RULES

(uv)^{'} = u^{'} v + u v^{'}, (\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} (v \neq = 0)

For $u, v$ differentiable at the point.

CHAIN RULE

\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)

For $g$ differentiable at $x$ and $f$ differentiable at $g (x)$ .

$f (x)$	$f^{'} (x)$	Domain note
$x^{n}$	$n x^{n - 1}$	positive integer $n$ : all $x$ ; negative integer $n$ : $x \neq = 0$ ; other real $n$ : $x > 0$
$e^{x}$	$e^{x}$	all $x$
$a^{x}$	$a^{x} ln a$	$a > 0$
$ln x$	$1/ x$	$x > 0$
$ln ∣ x ∣$	$1/ x$	$x \neq = 0$
$sin x$ , $cos x$	$cos x$ , $- sin x$	all $x$
$tan x$	$sec^{2} x$	$cos x \neq = 0$
$sin^{- 1} x$	$1/ 1 - x^{2}$	$∣ x ∣ < 1$
$cos^{- 1} x$	$- 1/ 1 - x^{2}$	$∣ x ∣ < 1$
$tan^{- 1} x$	$1/ (1 + x^{2})$	all $x$

LOGARITHMIC DIFFERENTIATION

\frac{d}{d x} u (x)^{v (x)} = u^{v} (v^{'} ln u + \frac{v u ^{'}}{u})

For $u > 0$ and $u, v$ differentiable. For example $\frac{d}{d x} x^{x} = x^{x} (1 + ln x)$ for $x > 0$ .

EXAM TIP

For $∣ g (x) ∣$ with $g$ differentiable, differentiability can fail only where $g (x) = 0$ . If $g (a) = 0$ and $g^{'} (a) \neq = 0$ , $∣ g ∣$ has a corner at $a$ (slopes $\pm g^{'} (a)$ ). If $g (a) = 0$ and $g^{'} (a) = 0$ (for a polynomial, a repeated root), $∣ g ∣$ is still differentiable at $a$ with derivative $0$ , because $∣ g (a + h) ∣/∣ h ∣ \to ∣ g^{'} (a) ∣ = 0$ .

Rolle's and the mean value theorems

ROLLE'S THEOREM

f continuous on [a, b], differentiable on (a, b), f (a) = f (b) \Rightarrow \exists c \in (a, b) : f^{'} (c) = 0

All three conditions are needed. Typical use: if $f$ is continuous on the closed interval between two roots and differentiable inside it, then $f^{'}$ has a root strictly between them.

LAGRANGE'S MEAN VALUE THEOREM

f continuous on [a, b], differentiable on (a, b) \Rightarrow \exists c \in (a, b) : f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}

Some point has tangent slope equal to the average slope over $[a, b]$ . Rolle's theorem is the special case $f (a) = f (b)$ .

CAUCHY'S MEAN VALUE THEOREM

f, g continuous on [a, b], differentiable on (a, b), g^{'} (x) \neq = 0 on (a, b) \Rightarrow \exists c \in (a, b) : \frac{f ^{'} ( c )}{g ^{'} ( c )} = \frac{f ( b ) - f ( a )}{g ( b ) - g ( a )}

The condition $g^{'} \neq = 0$ ensures $g (b) \neq = g (a)$ (by Rolle's theorem). Taking $g (x) = x$ gives Lagrange's theorem.

Consequences and how to use them

If $f$ is differentiable on an interval and $f^{'} (x) = 0$ for all $x$ in it, then $f$ is constant there.
If $f$ is differentiable on an interval and $f^{'} (x) > 0$ there, $f$ is strictly increasing there; if $f^{'} (x) < 0$ , strictly decreasing. The converse fails: $x^{3}$ is strictly increasing but $f^{'} (0) = 0$ .
To find $c$ , solve $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ and keep only solutions strictly inside $(a, b)$ .
The conditions are sufficient, not necessary: a function can have such a $c$ even if a condition fails, but failing a condition removes the guarantee.

WORKED EXAMPLE

Verify Lagrange's mean value theorem for $f (x) = x^{2} + 2 x$ on $[0, 4]$ and find $c$ .

Solution

$f$ is a polynomial, so it is continuous on $[0, 4]$ and differentiable on $(0, 4)$ .
Average slope: $\frac{f ( 4 ) - f ( 0 )}{4 - 0} = \frac{24 - 0}{4} = 6$ .
$f^{'} (c) = 2 c + 2 = 6$ gives $c = 2$ , which lies in $(0, 4)$ .

Answer: $c = 2$

EXAM TIP

Rolle's theorem can fail when differentiability fails at one interior point. For $f (x) = ∣ x ∣$ on $[- 1, 1]$ we have $f (- 1) = f (1)$ , yet $f^{'} (x) \neq = 0$ wherever it exists, because the only candidate $x = 0$ is a corner.

PREVIOUS-YEAR QUESTIONS

6 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2025 · GATE DA 2024

GATE 2025 DA Q.24

GATE 2025 · DAQ.24 · 1 mark · MSQ

Consider two functions $f : R \to R$ and $g : R \to (1, \infty)$ . Both functions are differentiable at a point $c$ . Which of the following functions is/are ALWAYS differentiable at $c$ ? The symbol $\cdot$ denotes product and the symbol $\circ$ denotes composition of functions.

(A)
$f \pm g$
(B)
$f \cdot g$
(C)
$\frac{f}{g}$
(D)
$f \circ g + g \circ f$

Answer and solution

GATE 2025 DA Q.24: answer and solution

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GATE 2025 DA Q.32

GATE 2025 · DAQ.32 · 1 mark · NAT

$t \to + \infty lim t^{2} + t - t =$ (Round off to one decimal place)

Answer and solution

GATE 2025 DA Q.32: answer and solution

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GATE 2025 DA Q.59

GATE 2025 · DAQ.59 · 2 marks · NAT

Let $f : R \to R$ be such that $∣ f (x) - f (y) ∣ \leq (x - y)^{2}$ for all $x, y \in R$ . Then $f (1) - f (0) =$ ______ (Answer in integer)

Answer and solution

GATE 2025 DA Q.59: answer and solution

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GATE 2024 DA Q.33

GATE 2024 · DAQ.33 · 1 mark · NAT

Let $f : R \to R$ be the function $f (x) = \frac{1}{1 + e ^{- x}}$ . The value of the derivative of $f$ at $x$ where $f (x) = 0.4$ is ______ (rounded off to two decimal places). Note: $R$ denotes the set of real numbers.

Answer and solution

GATE 2024 DA Q.33: answer and solution

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GATE 2024 DA Q.37

GATE 2024 · DAQ.37 · 2 marks · MCQ

Let $f : R \to R$ be a function. Note: $R$ denotes the set of real numbers.

f (x) = ⎩ ⎨ ⎧ - x, a x^{2} + b x + c, x, if x < - 2 if x \in [- 2, 2] if x > 2

Which ONE of the following choices gives the values of $a, b, c$ that make the function $f$ continuous and differentiable?

(A)
$a = \frac{1}{4}, b = 0, c = 1$
(B)
$a = \frac{1}{2}, b = 0, c = 0$
(C)
$a = 0, b = 0, c = 0$
(D)
$a = 1, b = 1, c = - 4$

Answer and solution

GATE 2024 DA Q.37: answer and solution

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GATE 2024 DA Q.60

GATE 2024 · DAQ.60 · 2 marks · NAT

Evaluate the following limit:

x \to 0 lim \frac{ln ( ( x ^{2} + 1 ) cos x )}{x ^{2}} = .

Answer and solution

GATE 2024 DA Q.60: answer and solution

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PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

Evaluate $x \to 0 lim \frac{e ^{3 x} - 1 - 3 x}{x ^{2}}$ . Give the value as a decimal.

Answer and solution

Answer: $4.5$ (accepted range $\pm 0.01$ )

At $x = 0$ the form is $0/0$ , so apply L'Hopital's rule.
Differentiate: numerator $3 e^{3 x} - 3$ , denominator $2 x$ . This is still $0/0$ .
Differentiate again: numerator $9 e^{3 x}$ , denominator $2$ .
At $x = 0$ this is $9/2 = 4.5$ .

PRACTICE 22 marks · MCQ

The number of points in $R$ at which $f (x) = ∣ x^{2} - 4 x + 3 ∣$ is not differentiable is

(A)
0
(B)
1
(C)
2
(D)
3

Answer and solution

Answer: (C)

$∣ g ∣$ can fail to be differentiable only where $g = 0$ .
$g (x) = x^{2} - 4 x + 3 = (x - 1) (x - 3)$ , so the zeros are $x = 1$ and $x = 3$ .
At each, $g^{'} (x) = 2 x - 4$ equals $- 2$ and $2$ , which are nonzero, so $∣ g ∣$ has a corner at both.
Hence exactly 2 points.

PRACTICE 32 marks · MSQ

Let $f (x) = x^{2} sin (1/ x)$ for $x \neq = 0$ and $f (0) = 0$ . Which of the following statements are TRUE?

(A)
$f$ is continuous at $x = 0$
(B)
$f$ is differentiable at $x = 0$
(C)
$f^{'}$ is continuous at $x = 0$
(D)
$f^{'} (0) = 1$

Answer and solution

Answer: (A), (B)

$∣ f (x) ∣ \leq x^{2} \to 0$ , so $f (x) \to 0 = f (0)$ and $f$ is continuous: A is true.
$\frac{f ( h ) - f ( 0 )}{h} = h sin (1/ h)$ , and $∣ h sin (1/ h) ∣ \leq ∣ h ∣ \to 0$ , so $f^{'} (0) = 0$ exists: B is true and D is false.
For $x \neq = 0$ , $f^{'} (x) = 2 x sin (1/ x) - cos (1/ x)$ . The first term tends to $0$ but $cos (1/ x)$ oscillates, so $f^{'} (x)$ has no limit at $0$ and $f^{'}$ is not continuous there: C is false.

PRACTICE 41 mark · NAT

The function $f (x) = \frac{1 - cos 4 x}{x ^{2}}$ for $x \neq = 0$ , with $f (0) = k$ , is continuous at $x = 0$ . Find $k$ .

Answer and solution

Answer: $8$ (accepted range $\pm 0.01$ )

Continuity requires $k = lim_{x \to 0} \frac{1 - c o s 4 x}{x ^{2}}$ .
Write $\frac{1 - c o s 4 x}{x ^{2}} = 16 \cdot \frac{1 - c o s 4 x}{( 4 x ) ^{2}}$ .
As $x \to 0$ , $4 x \to 0$ and $\frac{1 - c o s u}{u ^{2}} \to \frac{1}{2}$ , so the limit is $16 \cdot \frac{1}{2} = 8$ .

PRACTICE 52 marks · NAT

For $f (x) = x^{3}$ on the interval $[1, 3]$ , Lagrange's mean value theorem gives a point $c \in (1, 3)$ with $f^{'} (c) = \frac{f ( 3 ) - f ( 1 )}{3 - 1}$ . Find $c$ , correct to two decimal places.

Answer and solution

Answer: $2.08$ (accepted range $\pm 0.01$ )

$f (3) - f (1) = 27 - 1 = 26$ , so the average slope is $26/2 = 13$ .
$f^{'} (c) = 3 c^{2} = 13$ , so $c^{2} = 13/3$ .
$c = 13/3 \approx 2.0817$ , which lies in $(1, 3)$ (the negative root does not).

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE DA

Functions of a single variable, limit, continuity and differentiability

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