JEE Advanced Maths Limits Previous Year Questions with Solutions

Limits JEE Advanced previous year questions with solutions are given on this page. These solutions are designed by our team of experts. Limits is one of the most important topics of exams like the JEE Main and JEE Advanced. Besides being simple, this topic is scoring too. Hence, those who are aiming for exams like JEE Advanced must master this topic in order to remain competitive. Students are advised to learn and revise these solutions so that they can perform well in the upcoming JEE Main and JEE Advanced exams.

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Question 1: Let [t] denote the greatest integer ≤t. If for some λ∈R -{0, 1}, lim_{x→ 0}|(1-x+|x|)/(λ-x+[x])| = L, then L is equal to

(a) 1

(b) 2

(c) 1/2

(d) 0

Answer: b

Solution:

Given lim_{x→ 0}|(1-x+|x|)/(λ-x+[x])| = L

LHL = lim_{h→ 0}|(1+h+h)/(λ-h-1)| = |1/(λ-1)|

RHL = lim_{h→ 0} |(1-h+h)/(λ+h+0)| = |1/λ|

Given that limit exists.

So LHL = RHL

⇒ |λ-1| = |λ|

⇒ λ = 1/2

L = |1/λ| = 2

Hence option b is the answer.

Question 2: Let [x] be the greatest integer less than or equal to x. Then at which of the following point(s) the function f(x) = x cos (π(x + [x])) is discontinuous?

(a) x = 2

(b) x = 0

(c) x = 1

(d) x = -1

Answer: a, c, d

Solution:

Given f(x) = x cos (π(x + [x]))

At x = 2

lim_{x→ 2-}x cos (π(x + [x])) = 2 cos (π+2π)

= 2 cos 3π

= -2

lim_{x→ 2+}x cos (π(x + [x])) = 2 cos (2π+2π)

= 2 cos 4π

= 2

LHL ≠ RHL

So f(x) is discontinuous at x = 2.

At x = 0

lim_{x→ 0-}x cos (π(x + [x])) = 0 cos (-π+0)

= 0

lim_{x→ 0+}x cos (π(x + [x])) = 0

LHL = RHL

So f(x) is continuous at x = 0.

At x = 1

lim_{x→ 1-}x cos (π(x + [x])) = cos (π)

= -1

lim_{x→ 1+}x cos (π(x + [x])) = cos 2π

= 1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

At x = -1

lim_{x→ -1-}x cos (π(x + [x])) = -cos (-3π)

= 1

lim_{x→ -1+}x cos (π(x + [x])) = -cos 2π

= -1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

The function is discontinuous at x = 2, 1, -1

Question 3:

(a) 0

(b) 2

(c) 4

(d) 1

Answer: b

Solution:

lim_x→0 (x cot (4x))/sin²x cot² (2x) = lim_x→0 (x tan² 2x)/sin²x tan 4x

= lim_{x→ 0} (x/sin x)² (tan 2x/2x)² (4x/tan 4x)(4/2²)

= 1

Question 4: lim_{x→ 0} (1 – cos 2x)(3 + cos x)/x tan 4x is equal to

(a) 1/2

(b) 2

(c) 4

(d) 3

Answer: b

Solution:

Multiply and divide by 4x in the given expression, we get

=> lim_{x→ 0} [(1 – cos 2x)](3 + cos x) 4x/ 4x² tan 4x)

= lim_{x→ 0} [[2 sin²x/x²] [(3+cos x)/4] (4x/tan 4x)]

= 2 lim_{x→ 0} [sin²x/x²] lim_{x→ 0} [(3+cos x)/4] [1/lim_{x→ 0} (tan 4x)/4x]

= 2×(4/4)×1

= 2

Hence option b is the answer.

Question 5:

(a) 3

(b) 2

(c) 3/2

(d) 5/4

Answer: c

Solution:

(applying L Hospitals’ rule)

again applying L Hospitals’ rule and solving we get

= 1 + ½

= 3/2

Hence option c is the answer.

Question 6: lim_{x→ 0} sin (π cos²x)/x² is equal to

(a) -π

(b) π

(c) π/2

(d) 1

Answer: b

Solution:

lim_{x→ 0} sin (π cos²x)/x² = lim_{x→ 0} sin [π(1-sin²x)]/x²

= lim_{x→ 0} sin (π – π sin²x)/x²

= lim_{x→ 0} [sin (π sin²x)/(π sin²x)] (π sin²x)/x²

= lim_{x→ 0} π (sin x/x)²

= π

Hence option b is the answer.

Question 7: If lim_{x→ 0} ((a-n)nx – tan x) sin nx/x² = 0, where n is non zero real number, then a is equal to

(a) 0

(b) (n+1)/n

(c) n

(d) n + 1/n

Answer: d

Solution:

lim_{x→ 0} ((a-n)nx – tan x) sin nx/x² = 0

⇒ lim_{x→ 0} (n sin nx/nx)[ (a-n)n – tan x/x] = 0

⇒ n. [(a-n)n-1] = 0

⇒ a = n + 1/n

Hence option d is the answer.

Question 8: If lim_{x→ 0} [(8/x⁸)( 1 – cos (x²/2) – cos (x²/4) + cos (x²/2) cos (x²/4))] = 2^-k. Then the value of k is

Solution:

Given lim_{x→ 0} [(8/x⁸)( 1 – cos (x²/2) – cos (x²/4) + cos (x²/2) cos (x²/4))] = 2^-k

⇒ lim_{x→ 0} [(8/x⁸)((1 – cos (x²/2)) – (cos (x²/4) (1 – cos (x²/2)))] = 2^-k

Taking (1 – cos (x²/2) as common

⇒ lim_{x→ 0} (8/x⁸)(1 – cos (x²/2)[ 1 – cos (x²/4)] = 2^-k

(Use 1- cos 2x = 2 sin² x)

⇒ lim_{x→ 0}(8/x⁸) 2 sin² (x²/4) 2 sin² (x²/8) = 2^-k

⇒ lim_{x→ 0}(8/x⁸) 2 [sin (x²/4)]² 2 [sin (x²/8)]² = 2^-k

⇒ lim_{x→ 0} (32/x⁸) [sin (x²/4)]² [sin (x²/8)]² = 2^-k

⇒ lim_{x→ 0} (32) [(sin (x²/4))²] [sin (x²/8)]² /[(16x⁴)/16][(64x⁴)/64 = 2^-k

⇒ lim_{x→ 0} (32/16× 64) = 2^-k

⇒ 1/32 = 2^-k

⇒ 1/2⁵ = 1/2^k

⇒ k = 5

Question 9: lim_{x→ 2} (3^x + 3^3-x – 12)/(3^-x/2-3^1-x) is equal to

(a) 36

(b) 27

(c) 1

(d) -1

Answer: a

Solution:

lim_{x→ 2} (3^x + 3^3-x – 12)/(3^-x/2-3^1-x) = lim_{x→ 2} (3^2x – 12. 3^x + 27)/(3^x/2-3)

= lim_{x→ 2} (3^x – 9) (3^x – 3)/(3^x/2-3)

= lim_{x→ 2} (3^x/2 + 3)(3^x/2 – 3)(3^x – 3)/(3^x/2 – 3)

= 36

Hence option a is the answer.

Question 10: Let e denote the base of natural logaithm. The value of the real number a for which the right hand limit lim_x→0+ ((1-x)^1/x – e^-1)/x^a is equal to a non zero number is

(a) 0

(b) 1

(c) 2

(d) -2

Answer: b

Solution:

The value of a = 1

Hence option b is the answer.

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