JEE Mathematics Limits MCQs Set 18

Multiple Choice Questions (MCQs) for JEE Mathematics: Limits

Explore reliable objective questions for Limits tailored for JEE learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Limits MCQs for JEE Mathematics

View or download the dedicated Limits MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: If \( \lim_{x\to0} kx\,\operatorname{cosec}x=\lim_{x\to0}x\,\operatorname{cosec}kx \), then k =
(a) 1
(b) −1
(c) ±1
(d) ±2
Show Answer & Explanation

Answer: (c) ±1

Explanation:
1. The left limit is k, since x/sin x → 1.
2. The right limit is x/sin kx → 1/k.
3. So k = 1/k, i.e. k² = 1 and k = ±1.

Teacher's Note:
1. Use x/sin x → 1 on both sides.

Question 2: Let \( f(x)=\alpha(x)\beta(x)\gamma(x) \) for all real x, where α, β, γ are differentiable. If f′(2) = 18f(2), α′(2) = 3α(2), β′(2) = −4β(2) and γ′(2) = kγ(2), then k is
(a) 14
(b) 16
(c) 19
(d) None of these
Show Answer & Explanation

Answer: (c) 19

Explanation:
1. Logarithmic differentiation gives f′/f = α′/α + β′/β + γ′/γ.
2. At x = 2: 18 = 3 − 4 + k, so k = 19.

Teacher's Note:
1. Divide the product rule by f.

Question 3: \( \lim_{n\to\infty}\dfrac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{2n+3}} \) is equal to
(a) 5
(b) 3
(c) 1
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. The denominator is dominated by 32n+3 = 27 · 9n.
2. The numerator grows like 5 · 5n at most, which is much smaller than 9n.
3. So the limit is 0.

Teacher's Note:
1. Compare the largest exponential in numerator and denominator.

Question 4: \( \lim_{x\to0}\left[\min(y^2-4y+11)\dfrac{\sin x}{x}\right] \), where [·] is the greatest integer function, is
(a) 5
(b) 6
(c) 7
(d) does not exist
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. y² − 4y + 11 = (y − 2)² + 7 has minimum 7.
2. The expression is [7 sin x / x]. Near 0 (x ≠ 0), sin x / x is slightly less than 1, so 7 sin x / x lies just below 7 and above 6.
3. Hence the greatest integer is 6.

Teacher's Note:
1. Find the minimum first, then see whether sin x / x approaches 1 from below.

Question 5: \( \lim_{x\to0}\left[\dfrac{\sin(\operatorname{sgn}x)}{\operatorname{sgn}x}\right] \), where [·] is the greatest integer function, is
(a) 0
(b) 1
(c) −1
(d) does not exist
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. For x > 0, sgn x = 1 and the value is sin 1 ≈ 0.84.
2. For x < 0, sgn x = −1 and the value is sin(−1)/(−1) = sin 1 ≈ 0.84.
3. The greatest integer of 0.84 is 0 on both sides, so the limit is 0.

Teacher's Note:
1. sin t / t is even in t.

Question 6: For \( f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\dots+\dfrac{x^2}{2}+x+1 \), f′(1) = m f′(0) where m is equal to
(a) 50
(b) 0
(c) 100
(d) 200
Show Answer & Explanation

Answer: (c) 100

Explanation:
1. f′(x) = x99 + x98 + ... + x + 1.
2. f′(1) = 100 and f′(0) = 1.
3. So m = 100.

Teacher's Note:
1. Differentiate term by term.

Question 7: If \( y=(1+x^{1/4})(1+x^{1/2})(1-x^{1/4}) \), then dy/dx is equal to
(a) 1
(b) −1
(c) x
(d) \( \sqrt{x} \)
Show Answer & Explanation

Answer: (b) −1

Explanation:
1. (1 + x1/4)(1 − x1/4) = 1 − x1/2.
2. Then y = (1 − x1/2)(1 + x1/2) = 1 − x.
3. So dy/dx = −1.

Teacher's Note:
1. Use (a + b)(a − b) = a² − b² twice.

Question 8: If \( f(x)=\begin{cases}x^n\sin(1/x^2),&x\ne0\\0,&x=0\end{cases}\) (n ∈ I), then
(a) \( \lim_{x\to0}f(x) \) exists for n > 1
(b) \( \lim_{x\to0}f(x) \) exists for n < 0
(c) \( \lim_{x\to0}f(x) \) does not exist for any n
(d) \( \lim_{x\to0}f(x) \) cannot be determined
Show Answer & Explanation

Answer: (a) \( \lim_{x\to0}f(x) \) exists for n > 1

Explanation:
1. Since |sin(1/x²)| ≤ 1, |f(x)| ≤ |x|n, which tends to 0 for n > 0.
2. So the limit exists (and equals 0) for n > 1.
3. For n < 0, xn blows up while sin(1/x²) oscillates, so the limit does not exist.

Teacher's Note:
1. Use the squeeze theorem with |sin| ≤ 1.

Question 9: If x > 0 and g is a bounded function, then \( \lim_{n\to\infty}\dfrac{f(x)e^{nx}+g(x)}{e^{nx}+1} \) is
(a) 0
(b) f(x)
(c) g(x)
(d) None of these
Show Answer & Explanation

Answer: (b) f(x)

Explanation:
1. Divide numerator and denominator by enx.
2. Since x > 0, e−nx → 0, and g is bounded, so g(x)e−nx → 0.
3. The limit is f(x).

Teacher's Note:
1. Divide by the dominant term enx.

Question 10: The sequence \( \langle b_n\rangle \) satisfies \( b_{n+1}=\dfrac13\left(2b_n+\dfrac{125}{b_n^2}\right) \), bn ≠ 0. Then \( \lim b_n \) (assuming it exists and is positive) is
(a) 10
(b) 15
(c) 5
(d) 25
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. If the limit is L, then L = (2L + 125/L²)/3.
2. So 3L = 2L + 125/L², i.e. L³ = 125 and L = 5.

Teacher's Note:
1. Take limits on both sides of the recurrence.

Question 11: \( \lim_{x\to0}\dfrac{\sin[\cos x]}{1+[\cos x]} \) ([·] is the greatest integer function) is
(a) equal to 1
(b) equal to 0
(c) does not exist
(d) None of these
Show Answer & Explanation

Answer: (b) equal to 0

Explanation:
1. For x near 0 (x ≠ 0), 0 < cos x < 1, so [cos x] = 0.
2. The expression is sin 0 / (1 + 0) = 0.

Teacher's Note:
1. Find [cos x] near 0 first.

Question 12: If \( y=\left(1+\dfrac1x\right)\left(1+\dfrac2x\right)\cdots\left(1+\dfrac nx\right) \), x ≠ 0, then dy/dx at x = −1 is
(a) n!
(b) (n − 1)!
(c) (−1)n(n − 1)!
(d) (−1)n n!
Show Answer & Explanation

Answer: (c) (−1)n(n − 1)!

Explanation:
1. At x = −1 the first factor 1 + 1/x vanishes, so only the term where this factor is differentiated survives.
2. Its derivative is −1/x² = −1. The other factors at x = −1 are (1 − 2)(1 − 3)...(1 − n) = (−1)n−1(n − 1)!.
3. So dy/dx = −1 · (−1)n−1(n − 1)! = (−1)n(n − 1)!.

Teacher's Note:
1. Apply the product rule and note which terms contain the zero factor.

Question 13: A triangle has vertices P(a, 0), Q(0, b) and R(x, y) with R moving on y = x. If A is its area, then dA/dx is
(a) \( \frac{a-b}{2} \)
(b) \( \frac{a-b}{4} \)
(c) \( -\left(\frac{a+b}{2}\right) \)
(d) \( \frac{a+b}{4} \)
Show Answer & Explanation

Answer: (c) \( -\left(\frac{a+b}{2}\right) \)

Explanation:
1. With R = (x, x): A = (1/2)|a(b − x) + 0 + x(0 − b)| = (1/2)|ab − (a + b)x|.
2. Taking the expression inside as positive, A = (1/2)(ab − (a + b)x).
3. So dA/dx = −(a + b)/2.

Teacher's Note:
1. Use the determinant formula for area.

Question 14: The derivative of \( \left(\sqrt x+\dfrac1{\sqrt x}\right)^2 \) is
(a) \( \frac1{x^2} \)
(b) \( 1-\frac1{x^2} \)
(c) 1
(d) \( 1+\frac1{x^2} \)
Show Answer & Explanation

Answer: (b) \( 1-\frac1{x^2} \)

Explanation:
1. Expand: (√x + 1/√x)² = x + 2 + 1/x.
2. Differentiate: 1 − 1/x².

Teacher's Note:
1. Expand the square before differentiating.

Question 15: If f(x) = 2x² + 3x − 5, then f′(0) = m f′(−1), where m is
(a) −1
(b) −2
(c) −3
(d) −4
Show Answer & Explanation

Answer: (c) −3

Explanation:
1. f′(x) = 4x + 3, so f′(0) = 3 and f′(−1) = −1.
2. m = 3/(−1) = −3.

Teacher's Note:
1. Find f′ first.

Question 16: Let f(x) = x² − 1 for 0 < x < 2 and f(x) = 2x + 3 for 2 ≤ x < 3. The quadratic equation whose roots are \( \lim_{x\to2^-}f(x) \) and \( \lim_{x\to2^+}f(x) \) is
(a) x² − 6x + 9 = 0
(b) x² − 10x + 21 = 0
(c) x² − 14x + 49 = 0
(d) None of these
Show Answer & Explanation

Answer: (b) x² − 10x + 21 = 0

Explanation:
1. Left limit: 4 − 1 = 3. Right limit: 2(2) + 3 = 7.
2. Sum of roots 10 and product 21 give x² − 10x + 21 = 0.

Teacher's Note:
1. Evaluate each piece at x = 2.

Question 17: If x1 = 3 and \( x_{n+1}=\sqrt{2+x_n} \), n ≥ 1, then \( \lim x_n \) is equal to
(a) −1
(b) 2
(c) \( \sqrt5 \)
(d) 3
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. The sequence is decreasing and bounded below by 2, so it converges to L with L = √(2 + L).
2. L² = L + 2 gives (L − 2)(L + 1) = 0, and L ≥ 0 so L = 2.

Teacher's Note:
1. Solve L = √(2 + L).

Question 18: Let \( f(x)=\begin{cases}x\sin\frac1x+\sin\frac1{x^2},&x\ne0\\0,&x=0\end{cases}\). Then \( \lim_{x\to\infty}f(x) \) equals
(a) 0
(b) −1/2
(c) 1
(d) None of these
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. As x → ∞, x sin(1/x) = sin(1/x)/(1/x) → 1.
2. sin(1/x²) → 0.
3. So the limit is 1.

Teacher's Note:
1. Put t = 1/x in the first term.

Question 19: Let f(x) = x − [x] and \( g(x)=\lim_{n\to\infty}\dfrac{\{f(x)\}^{2n}-1}{\{f(x)\}^{2n}+1} \). Then g(x) equals
(a) 0
(b) 1
(c) −1
(d) None of these
Show Answer & Explanation

Answer: (c) −1

Explanation:
1. f(x) is the fractional part, so 0 ≤ f(x) < 1, and {f(x)} = f(x).
2. Then f(x)2n → 0, giving g(x) = (0 − 1)/(0 + 1) = −1 for every x.

Teacher's Note:
1. The fractional part is always below 1.

Question 20: \( \lim_{x\to0}|x|^{[\cos x]} \) is
(a) 1
(b) does not exist
(c) 0
(d) None of these
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. For x near 0, x ≠ 0, 0 < cos x < 1, so [cos x] = 0.
2. Then |x|0 = 1 and the limit is 1.

Teacher's Note:
1. Find [cos x] near 0 first.

Question 21: If \( \dfrac{d}{dx}\left(\dfrac{1+x^4+x^8}{1+x^2+x^4}\right)=ax^3+bx \), then
(a) a = 4, b = 2
(b) a = 4, b = −2
(c) a = −2, b = 4
(d) None of these
Show Answer & Explanation

Answer: (b) a = 4, b = −2

Explanation:
1. 1 + x⁴ + x⁸ = (1 + x² + x⁴)(1 − x² + x⁴), so the fraction is x⁴ − x² + 1.
2. Its derivative is 4x³ − 2x, so a = 4 and b = −2.

Teacher's Note:
1. Factorise 1 + x⁴ + x⁸.

Question 22: If \( f(x)=\left(\dfrac{\sin^mx}{\sin^nx}\right)^{m+n}\cdot\left(\dfrac{\sin^nx}{\sin^px}\right)^{n+p}\cdot\left(\dfrac{\sin^px}{\sin^mx}\right)^{p+m} \), then f′(x) is equal to
(a) 0
(b) 1
(c) \( \cos^{m+n+p}x \)
(d) None of these
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. The exponent of sin x is (m − n)(m + n) + (n − p)(n + p) + (p − m)(p + m) = (m² − n²) + (n² − p²) + (p² − m²) = 0.
2. So f(x) = 1, a constant, and f′(x) = 0.

Teacher's Note:
1. Add the powers of sin x.

Question 23: Let \( f(x)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}} \), 1 < x < 26. Then f′(x) for 1 < x < 26 is
(a) 0
(b) \( \frac1{\sqrt{x-1}} \)
(c) \( 2\sqrt{x-1}-5 \)
(d) None of these
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. Put t = √(x − 1). Then x + 24 − 10t = t² + 25 − 10t = (t − 5)².
2. So f = t + |t − 5|. For 1 < x < 26, 0 < t < 5, so f = t + 5 − t = 5.
3. Hence f′(x) = 0.

Teacher's Note:
1. Write the inner expression as a perfect square.

Question 24: If \( y=\dfrac1{1+x^{\beta-\alpha}+x^{\gamma-\alpha}}+\dfrac1{1+x^{\alpha-\beta}+x^{\gamma-\beta}}+\dfrac1{1+x^{\alpha-\gamma}+x^{\beta-\gamma}} \), then dy/dx is equal to
(a) 0
(b) 1
(c) \( (\alpha+\beta+\gamma)x^{\alpha+\beta+\gamma-1} \)
(d) None of these
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. Put p = xα, q = xβ, r = xγ. The first term is 1/(1 + q/p + r/p) = p/(p + q + r).
2. Similarly the others are q/(p + q + r) and r/(p + q + r).
3. The sum is 1, a constant, so dy/dx = 0.

Teacher's Note:
1. Multiply numerator and denominator of each term by a suitable power of x.

Question 25: \( \lim_{n\to\infty}\dfrac{n^p\sin^2(n!)}{n+1} \), 0 < p < 1, is equal to
(a) 0
(b) ∞
(c) None of these
(d) 1
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. 0 ≤ sin²(n!) ≤ 1, so the expression lies between 0 and np/(n + 1).
2. Since p < 1, np/(n + 1) → 0, so the limit is 0 by the squeeze theorem.

Teacher's Note:
1. Bound sin² between 0 and 1.

Question 26: Let f(x) = x(−1)[1/x], x ≠ 0, where [x] is the greatest integer function. Then \( \lim_{x\to0}f(x) \) =
(a) does not exist
(b) 2
(c) 0
(d) −1
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. (−1)[1/x] is either 1 or −1, so |f(x)| = |x|.
2. As x → 0, |x| → 0, so the limit is 0.

Teacher's Note:
1. The factor (−1)[1/x] only changes the sign.

Question 27: \( \lim_{x\to0}\dfrac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)} \) is equal to
(a) 0
(b) −1/6
(c) 1/6
(d) does not exist
Show Answer & Explanation

Answer: (c) 1/6

Explanation:
1. With u = x⁴: sin u − u cos u ≈ u³/3 = x¹²/3, and x²⁰ is negligible.
2. The denominator is x⁴(e2u − 1 − 2u) ≈ x⁴ · 2u² = 2x¹².
3. The limit is (1/3)/2 = 1/6.

Teacher's Note:
1. Expand sin u, cos u and e2u up to the first non-vanishing order.

Free study material for Limits and Derivatives

Download Chapter MCQs: JEE Mathematics Limits

Download Multiple Choice Questions: Limits (JEE Mathematics)

Explore reliable practice questions for Limits tailored for JEE Mathematics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Concept Clarification for Limits

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Mathematics to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for JEE Mathematics

Follow up your worksheet practice by attempting the interactive online Mathematics MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Mathematics Limits MCQs Set 18?

You can get most exhaustive JEE Mathematics Limits MCQs Set 18 for free on StudiesToday.com. These MCQs for JEE Mathematics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics JEE material?

Yes, our JEE Mathematics Limits MCQs Set 18 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in JEE exams?

By solving our JEE Mathematics Limits MCQs Set 18, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for JEE Mathematics Limits MCQs Set 18?

Yes, Mathematics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Mathematics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Mathematics Limits MCQs Set 18 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.