Limits and Continuity PYQs for GATE CS

Solve 4+ Limits and Continuity previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Limits and Continuity

Chapter Journey: Limits & Continuity

Evaluation of Indeterminate Limits
L'Hôpital's Rule, Standard Limits, Series Expansion. The core engine.

Continuity of Piecewise Functions
Matching left-hand and right-hand limits. Finding unknown constants.

Discontinuities of Digit-Defined Functions
Advanced logic-based continuity checks. Rare but high-difficulty.

Goal: Master the algebraic manipulation of limits first. Everything else depends on it.

The Hero Concept: Why Indeterminate Forms?

What is an Indeterminate Form?

A limit $lim_{x \to a} \frac{f ( x )}{g ( x )}$ is indeterminate if direct substitution yields:

\frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, \infty - \infty, 1^{\infty}, 0^{0}, \infty^{0}

Numerator vs Denominator approaching zero

If numerator approaches zero faster, limit is $0$ .
If denominator approaches zero faster, limit is $\infty$ .
If comparable, limit is a finite non-zero number.

Limits and Continuity: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Engineering Mathematics · 2026_Set2 NAT

Consider a function

f : (0, 1) \to {0, 1}

defined as follows.

For a real number

r \in (0, 1)

f (r) = 1

if the second digit after the decimal point in

r

is one of the four digits 2, 3, 6 and 7. Otherwise,

f (r)

is equal to 0.

The number of points in

(0, 1)

at which

f

is discontinuous is ___________. (answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a digit-defined function continuity question. The function value depends on the second decimal digit of $r$ . Discontinuities occur only at points where the second digit changes AND the function value changes.

Step 1: Understand the second digit $d_{2} (r)$ . For $r \in (0, 1)$ , write $r = 0. d_{1} d_{2} d_{3} \dots$ . The second digit $d_{2}$ is constant on each interval $[k /100, (k + 1) /100)$ for $k = 0, 1, \dots, 99$ . Specifically, $d_{2} (r) = k mod 10$ for $r \in [k /100, (k + 1) /100)$ .

Step 2: Identify where $f$ can be discontinuous. Since $f$ is constant on the interior of each interval $[k /100, (k + 1) /100)$ , it is continuous there. Discontinuities can only occur at the boundary points $x = k /100$ for $k = 1, 2, \dots, 99$ .

Step 3: Check continuity at each boundary $x = k /100$ . At such a point:

$f (x)$ depends on $d_{2} (x) = k mod 10$ .
Left limit: as $r \to x^{-}$ , $r$ is in $[(k - 1) /100, k /100)$ , so $d_{2} (r) = (k - 1) mod 10$ .
Right limit: as $r \to x^{+}$ , $r$ is in $[k /100, (k + 1) /100)$ , so $d_{2} (r) = k mod 10$ .

Step 4: $f$ is discontinuous at $x = k /100$ if and only if $f ((k - 1) mod 10) \neq = f (k mod 10)$ . Let's compute $f (d)$ for $d = 0, 1, \dots, 9$ :

$f (0) = 0, f (1) = 0, f (2) = 1, f (3) = 1, f (4) = 0, f (5) = 0, f (6) = 1, f (7) = 1, f (8) = 0, f (9) = 0$ .

Step 5: Find transitions where $f$ changes value:

$d = 1 \to 2$ : $f (1) = 0 \to f (2) = 1$ . Change! Occurs at $k \equiv 2 (mod 10)$ : $k = 2, 12, 22, \dots, 92$ (10 values).
$d = 3 \to 4$ : $f (3) = 1 \to f (4) = 0$ . Change! Occurs at $k \equiv 4 (mod 10)$ : $k = 4, 14, 24, \dots, 94$ (10 values).
$d = 5 \to 6$ : $f (5) = 0 \to f (6) = 1$ . Change! Occurs at $k \equiv 6 (mod 10)$ : $k = 6, 16, 26, \dots, 96$ (10 values).
$d = 7 \to 8$ : $f (7) = 1 \to f (8) = 0$ . Change! Occurs at $k \equiv 8 (mod 10)$ : $k = 8, 18, 28, \dots, 98$ (10 values).

Other transitions ( $0 \to 1$ , $2 \to 3$ , $4 \to 5$ , $6 \to 7$ , $8 \to 9$ , $9 \to 0$ ) do not change $f$ , so no discontinuity.

Step 6: Total discontinuities = $4 \times 10 = 40$ .

Answer: 40

Question 2 · Engineering Mathematics · 2026_Set1 NAT

Consider the function

f : R \to R

defined as follows:

f (x) = ⎩ ⎨ ⎧ c_{1} e^{x} - c_{2} lo g_{e} (\frac{1}{x}), 3, if x > 0 otherwise

where

c_{1}, c_{2} \in R

.

If

f

is continuous at

x = 0

, then

c_{1} + c_{2} =

_________. (answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a piecewise continuity question with a hidden singularity. The function has different rules for $x > 0$ and $x \leq 0$ . For continuity at $x = 0$ , the right-hand limit must equal the left-hand limit and the function value. The term $ln (1/ x)$ blows up as $x \to 0^{+}$ , which forces its coefficient to be zero.

Step 1: Identify the junction point. The function changes rule at $x = 0$ . For continuity at $x = 0$ , we need:

$x \to 0^{-} lim f (x) = f (0) = x \to 0^{+} lim f (x)$

Step 2: Compute the left-hand limit and function value. For $x \leq 0$ , $f (x) = 3$ . So:

$x \to 0^{-} lim f (x) = 3, f (0) = 3$

Step 3: Compute the right-hand limit. For $x > 0$ :

$f (x) = c_{1} e^{x} - c_{2} ln (1/ x) = c_{1} e^{x} + c_{2} ln (x)$

As $x \to 0^{+}$ :

$e^{x} \to 1$ , so $c_{1} e^{x} \to c_{1}$ .
$ln (x) \to - \infty$ .

Step 4: For $lim_{x \to 0^{+}} f (x)$ to be finite (and equal to 3), the $ln (x)$ term must vanish. This requires $c_{2} = 0$ .

Step 5: With $c_{2} = 0$ :

$x \to 0^{+} lim f (x) = x \to 0^{+} lim c_{1} e^{x} = c_{1} \cdot 1 = c_{1}$

For continuity, $c_{1} = 3$ .

Step 6: Therefore, $c_{1} = 3$ , $c_{2} = 0$ , and $c_{1} + c_{2} = 3 + 0 = 3$ .

Answer: 3

Question 3 · Engineering Mathematics · 2022 NAT

The value of the following limit is _____________.

$x \to 0^{+} lim \frac{x}{1 - e ^{2 x}}$

Correct Answer:

-0.5

Step-by-Step Solution

Key idea: This is an indeterminate limit of the form $0/0$ involving a square root and an exponential. The key is to substitute $t = x$ to simplify, then use the standard limit $lim_{u \to 0} \frac{e ^{u} - 1}{u} = 1$ .

Step 1: Substitute $t = x$ . As $x \to 0^{+}$ , $t \to 0^{+}$ . The limit becomes:

$t \to 0^{+} lim \frac{t}{1 - e ^{2 t}}$

Step 2: This is still $0/0$ form. Rewrite the denominator:

$\frac{t}{1 - e ^{2 t}} = \frac{t}{- ( e ^{2 t} - 1 )} = - \frac{t}{e ^{2 t} - 1}$

Step 3: Multiply and divide by 2 to match the standard limit form:

$- \frac{t}{e ^{2 t} - 1} = - \frac{1}{2} \cdot \frac{2 t}{e ^{2 t} - 1}$

Step 4: As $t \to 0^{+}$ , let $u = 2 t \to 0^{+}$ . Then:

$t \to 0^{+} lim \frac{2 t}{e ^{2 t} - 1} = u \to 0^{+} lim \frac{u}{e ^{u} - 1} = 1$

Step 5: Therefore:

$t \to 0^{+} lim \frac{t}{1 - e ^{2 t}} = - \frac{1}{2} \cdot 1 = - \frac{1}{2} = - 0.5$

Answer: -0.5

Question 4 · Engineering Mathematics · 2021_Set1 NAT

Consider the following expression.

x \to - 3 lim \frac{2 x + 22 - 4}{x + 3}

The value of the above expression (rounded to 2 decimal places) is __________.

Correct Answer:

0.25

Step-by-Step Solution

Key idea: This is an indeterminate limit of the form $0/0$ involving a square root. The standard technique is to rationalize the numerator by multiplying by the conjugate.

Step 1: Check the form. At $x = - 3$ :

Numerator: $2 (- 3) + 22 - 4 = 16 - 4 = 4 - 4 = 0$ .
Denominator: $- 3 + 3 = 0$ .

So it's $0/0$ form.

Step 2: Rationalize the numerator. Multiply numerator and denominator by the conjugate $2 x + 22 + 4$ :

$\frac{2 x + 22 - 4}{x + 3} \cdot \frac{2 x + 22 + 4}{2 x + 22 + 4}$

Step 3: Simplify the numerator using $(a - b) (a + b) = a^{2} - b^{2}$ :

$(2 x + 22)^{2} - 4^{2} = (2 x + 22) - 16 = 2 x + 6 = 2 (x + 3)$

Step 4: The expression becomes:

$\frac{2 ( x + 3 )}{( x + 3 ) ( 2 x + 22 + 4 )}$

Step 5: Cancel the common factor $(x + 3)$ (valid since $x \to - 3$ but $x \neq = - 3$ ):

$\frac{2}{2 x + 22 + 4}$

Step 6: Now substitute $x = - 3$ :

$\frac{2}{2 ( - 3 ) + 22 + 4} = \frac{2}{16 + 4} = \frac{2}{4 + 4} = \frac{2}{8} = \frac{1}{4} = 0.25$

Answer: 0.25

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Limits and Continuity PYQs for GATE CS

Chapter Roadmap: Limits and Continuity

Chapter Journey: Limits & Continuity

The Hero Concept: Why Indeterminate Forms?

What is an Indeterminate Form?

Limits and Continuity: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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