JEE Mathematics Limits MCQs Set 17

Answer: (c) 8

Explanation:
1. Put \( x = \pi/4 + h \). Both numerator and denominator vanish, so apply L'Hospital's rule.
2. Numerator derivative: \( -3\cot^{2}x \csc^{2}x - \sec^{2}x = -3(1)(2) - 2 = -8 \) at \( x = \pi/4 \). Denominator derivative: \( -\sin(x + \pi/4) = -1 \).
3. The limit is \( (-8)/(-1) = 8 \).

Teacher's Note:
1. The form is 0/0 at \( x = \pi/4 \), so differentiate top and bottom.

Answer: (a) does not exist

Explanation:
1. Right limit \( (x \to 0^{+}, [x] = 0) \): \( \tan(\pi \sin^{2}x)/x^{2} \to \pi \) and \( (|x|)^{2}/x^{2} = 1 \), so the limit is \( \pi + 1 \).
2. Left limit \( (x \to 0^{-}, [x] = -1) \): \( |x| - \sin x[x] = -x + \sin x = O(x^{3}) \), so the second term is \( O(x^{6})/x^{2} \to 0 \); the limit is π.
3. The two one-sided limits differ, so the limit does not exist.

Teacher's Note:
1. Evaluate the left and right limits separately when a greatest integer function is present.

Answer: (a) −sin 1

Explanation:
1. For \( x \) slightly less than 0: \( [x] = -1, |x| = -x \), and \( \sin[x] = -\sin 1 \).
2. The expression becomes \( x(-1 - x)(-\sin 1)/(-x) \). Since \( x/(-x) = -1 \), this equals \( (-1)(-1 - x)(-\sin 1) = -(1 + x) \sin 1 \).
3. As \( x \to 0^{-} \) this tends to \( -\sin 1 \).

Teacher's Note:
1. Replace \( [x] \) and \( |x| \) by their values on the left of 0 and cancel \( x \).

Answer: (b) 1

Explanation:
1. As \( x \to 0 \): \( x \cot 4x \to 1/4 \) (since \( \cot 4x \approx 1/(4x) \)).
2. \( \sin^{2}x \cot^{2}2x \to x^{2} \cdot 1/(4x^{2}) = 1/4 \).
3. The ratio is \( (1/4)/(1/4) = 1 \).

Teacher's Note:
1. Replace \( \cot \) by 1/(angle) for small angles.

Answer: (c) exists and equals \( \frac{1}{4\sqrt2} \)

Explanation:
1. Let \( u = y^{4} \) and \( g(u) = \sqrt (1 + \sqrt{1 + u}) \). The limit is \( g'(0) \).
2. \( g'(u) = 1/(2\sqrt (1 + \sqrt{1 + u})) \cdot 1/(2\sqrt{1 + u}) \), so \( g'(0) = 1/(2\sqrt{2} \cdot 2) = 1/(4\sqrt{2}) \).

Teacher's Note:
1. A limit of the form \( (g(u) - g(0))/u \) is just \( g'(0) \).

Answer: (d) \( \sqrt{2/\pi} \)

Explanation:
1. Near \( x = 1, \sin^{-1}x \approx \pi/2 - \sqrt (2(1 - x)) \). Let \( t = \sqrt{1 - x} \), so \( 2 \sin^{-1}x \approx \pi - 2\sqrt{2} t \).
2. \( \sqrt{\pi - 2\sqrt{2} t} \approx \sqrt \pi - \sqrt{2} t/\sqrt \pi \).
3. The ratio tends to \( \sqrt{2}/\sqrt \pi = \sqrt{2/\pi} \).

Teacher's Note:
1. Use \( \sin^{-1}x = \pi/2 - \cos^{-1}x \) and \( \cos^{-1}x \approx \sqrt (2(1 - x)) \) near \( x = 1 \).

Answer: (b) −1/2 and −1

Explanation:
1. For small a: ∛(1 + a) − 1 ≈ a/3, √(1 + a) − 1 ≈ a/2, ⁶√(1 + a) − 1 ≈ a/6.
2. Dividing the equation by a and letting \( a \to 0 \): \( x^{2}/3 + x/2 + 1/6 = 0 \), i.e. \( 2x^{2} + 3x + 1 = 0 \).
3. The roots are \( x = -1/2 \) and \( x = -1 \).

Teacher's Note:
1. Use (1 + a)^k − 1 ≈ ka for small a.

Answer: (c) 3

Explanation:
1. \( \cos x - 1 \approx -x^{2}/2 \) and \( \cos x \) − eˣ ≈ \( -x \) (leading term).
2. The numerator behaves like \( x^{3}/2 \).
3. The limit is finite and non-zero only for \( n = 3 \) (it equals 1/2).

Teacher's Note:
1. Find the leading power of \( x \) in the numerator.

Answer: (a) −1

Explanation:
1. \( x^{4} \sin(1/x) = x^{3} - x/6 \) + ..., so the numerator is \( x^{3} + x^{2} - x/6 \) + ... For \( x \to -\infty, |x|^{3} = -x^{3} \).
2. The fraction tends to \( x^{3}/(-x^{3}) = -1 \). Writing \( x = -t \), it equals \( -1 + (t^{2} + t/6 + 1)/(1 + t^{3}) \), which is slightly greater than −1.
3. So the fraction lies just above −1 and its greatest integer is −1.

Teacher's Note:
1. Check from which side the value approaches the integer before taking [ ].

Answer: (b) 0

Explanation:
1. The numerator \( x - \sin x \) tends to 0.
2. The denominator \( x + \cos^{2}x \) tends to 1.
3. The limit is \( \sqrt{0/1} = 0 \).

Teacher's Note:
1. Not every limit near 0 is indeterminate; check the denominator first.

Answer: (d) 7

Explanation:
1. The numerator is about \( \betax^{3} \), so the denominator must also be of order \( x^{3} \): \( \alpha = 1 \) (otherwise the limit is 0).
2. Then \( x - \sin x \approx x^{3}/6 \), so the limit is \( 6\beta = 1 \) and \( \beta = 1/6 \).
3. \( 6(\alpha + \beta) = 6 + 1 = 7 \).

Teacher's Note:
1. Match the order of the numerator and the denominator to fix α.

Answer: (a) a = −3 and b = 9/2

Explanation:
1. \( \sin 3x = 3x - 27x^{3}/6 \) + ..., so \( x^{-3} \sin 3x = 3/x^{2} - 9/2 + O(x^{2}) \).
2. For the limit to exist the \( 1/x^{2} \) terms must cancel: \( 3 + a = 0 \), so \( a = -3 \).
3. The limit is \( -9/2 + b = 0 \), so \( b = 9/2 \).

Teacher's Note:
1. Expand \( \sin 3x \) and cancel the singular term.

Answer: (b) 0

Explanation:
1. Let \( t \) = xⁿ/eˣ → 0 as \( x \to \infty \). The numerator is 2^t − 3^t ≈ \( t \ln(2/3) \).
2. Dividing by \( x^{n} \) gives ln(2/3)/eˣ, which tends to 0.

Teacher's Note:
1. Exponential growth beats any power of \( x \).

Answer: (a) 1/5, 7/5

Explanation:
1. For the limit to be finite we need (x⁵)^c ~ \( x \), so \( c = 1/5 \).
2. (\( x^{5} + 7x^{4} \) + 2)^1/5 = x(1 + 7/x + 2/x⁵)^1/5 ≈ \( x + 7/5 \).
3. The limit is 7/5.

Teacher's Note:
1. Factor out \( x^{5} \) and expand with the binomial theorem.

Answer: (c) \( \text{If }\lim_{x\to c}(f(x)+g(x))\text{ and }\lim_{x\to c}f(x)\text{ exist, then }\lim_{x\to c}g(x)\text{ exists.} \)

Explanation:
1. \( (c) g = (f + g) - f \) is a difference of two functions whose limits exist, so \( \lim g \) exists.
2. \( (a) \) fails if \( \lim f = 0 \) (take \( f = 0, g = \sin(1/x), c = 0 \)). \( (b) \) and \( (d) \) fail: \( f = \sin(1/x), g = -\sin(1/x) \) has \( f + g = 0 \) but no limit for \( f \) or \( g \).

Teacher's Note:
1. A difference of convergent limits converges; products and sums alone do not force individual limits.

Answer: (d) \( (1-x)^{-1} \)

Explanation:
1. Multiply and divide by \( (1 - x) \): (\( 1 - x)(1 + x)(1 + x^{2})...(1 \) + x^2ⁿ) = 1 − x^2ⁿ⁺¹.
2. For \( |x| \) < 1, x^2ⁿ⁺¹ → 0, so the product tends to \( 1/(1 - x) \).

Teacher's Note:
1. The product telescopes by the identity \( (1 - a)(1 + a) = 1 - a^{2} \).

Answer: (a) \( \frac13 \)

Explanation:
1. Numerator = Σ \( k^{2}(n + 1 - k) = (n + 1)\cdot n(n + 1)(2n + 1)/6 - n^{2}(n + 1)^{2}/4 \approx n^{4}/3 - n^{4}/4 = n^{4}/12 \).
2. Denominator = \( n^{2}(n + 1)^{2}/4 \approx n^{4}/4 \).
3. The ratio tends to \( (1/12)/(1/4) = 1/3 \).

Teacher's Note:
1. Keep only the leading \( n^{4} \) terms in both sums.

Answer: (a) 1

Explanation:
1. x^−a log_a \( x \to 0 \), so \( \cot^{-1} \to \cot^{-1} 0 = \pi/2 \).
2. aˣ log_x \( a \to \infty \), so \( \sec^{-1} \to \pi/2 \).
3. The ratio tends to \( (\pi/2)/(\pi/2) = 1 \).

Teacher's Note:
1. Evaluate numerator and denominator separately; both tend to π/2.

Answer: (b) −16

Explanation:
1. As \( x \to 2^{-}, (\cos(2x - 4) - 33)/2 \to (1 - 33)/2 = -16 \).
2. For \( x \) < \( 2, |4x - 8| = -4(x - 2) \), so the right bound is \( -4x^{2} \to -16 \).
3. By the sandwich theorem the limit is −16.

Teacher's Note:
1. Squeeze theorem: find the limits of both bounds.

Answer: (a) 1

Explanation:
1. \( \sqrt{x + 1} - \sqrt{x} = 1/(\sqrt{x + 1} + \sqrt{x}) \to 0 \), so the numerator → \( \cot^{-1} 0 = \pi/2 \).
2. ((\( 2x + 1)/(x \) − 1))ˣ → ∞ (the base tends to 2), so \( \sec^{-1} \to \pi/2 \).
3. The limit is 1.

Teacher's Note:
1. Both inverse functions approach π/2.

Answer: (a) l = m

Explanation:
1. As \( x \to \infty \), eˣ dominates: \( f \approx \ln \) eˣ/ln e²ˣ = \( x/(2x) = 1/2 \), so \( l = 1/2 \).
2. As \( x \to -\infty \), eˣ → 0: \( f \approx \ln x^{2}/\ln x^{4} = 2 \ln|x|/(4 \ln|x|) = 1/2 \), so \( m = 1/2 \).
3. Hence \( l = m \).

Teacher's Note:
1. Identify the dominant term in each logarithm for each direction.

Answer: (d) \( -\frac{3\pi^2}{16} \)

Explanation:
1. Put \( t = \pi/2 - x \to 0 \). Then \( 1 - \sin x = 1 - \cos t \approx t^{2}/2, \cos x = \sin t \approx t \), and \( \pi - 2x = 2t \).
2. \( 8x^{3} - \pi^{3} = 8(x^{3} - (\pi/2)^{3}) \approx 8 \cdot 3(\pi/2)^{2} \cdot (-t) = -6\pi^{2}t \).
3. Numerator ≈ \( (t^{2}/2)(-6\pi^{2}t)(t) = -3\pi^{2}t^{4} \) and denominator = \( 16t^{4} \), so the limit is −3π²/16.

Teacher's Note:
1. Shift the variable so that the limit point becomes 0.

Answer: (b) \( \frac{1}{4\sqrt2} \)

Explanation:
1. \( \sqrt{x^{4} + 1} = x^{2}(1 + 1/(2x^{4}) + ...) \), so \( x^{2} + \sqrt{x^{4} + 1} = 2x^{2} + 1/(2x^{2}) \) + ...
2. \( \sqrt (2x^{2} + 1/(2x^{2})) = x\sqrt{2} \cdot (1 + 1/(8x^{4}) + ...) \approx x\sqrt{2} + \sqrt{2}/(8x^{3}) \).
3. Hence \( f(x) \approx x^{3} \cdot \sqrt{2}/(8x^{3}) = \sqrt{2}/8 = 1/(4\sqrt{2}) \).

Teacher's Note:
1. Expand each radical by the binomial series and subtract.

Answer: (c) −6

Explanation:
1. For \( x \) slightly less than \( -1/3, -1/x \) is slightly less than 3, so \( [-1/x] = 2 \).
2. The expression is \( 2/x \to 2/(-1/3) = -6 \).

Teacher's Note:
1. Find the integer value of the bracket near the limit point.

Answer: (d) \( \frac43 \)

Explanation:
1. Factor x^2/3: the bracket is x^2/3[(1 + 1/x)^2/3 − (1 − 1/x)^2/3].
2. The difference is about \( (2/3)(1/x) - (-(2/3)(1/x)) = 4/(3x) \).
3. The limit is x^1/3 · x^2/3 \( \cdot 4/(3x) = 4/3 \).

Teacher's Note:
1. Use the binomial expansion (1 ± h)^2/3 ≈ \( 1 \pm 2h/3 \).

Answer: (a) \( b^{1/a} \)

Explanation:
1. The expression is (1 + (b^1/n − 1)/a)ⁿ, of type \( 1^\infty \).
2. n(b^1/n − 1) → ln b, so the limit is \( \exp(\ln b / a) \) = b^1/a.

Teacher's Note:
1. For \( 1^\infty \) forms use e^{lim n(base − 1)}.

Answer: (c) 3

Explanation:
1. For small \( x \ne 0, f(x) = \tan x/x \) > 1, so \( [f(x)] = 1 \) and \( {f(x)} = f(x) - 1 \approx x^{2}/3 \).
2. The limit is (1 + x²)^3/x² → \( e^{3} \).
3. \( \ln(e^{3}) = 3 \).

Teacher's Note:
1. \( \tan x/x \) slightly exceeds 1 near 0, so \( [f] = 1 \).

Answer: (d) 2

Explanation:
1. (\( \sqrt{\tan} \) x)^√x = \( \exp((\sqrt{x}/2) \ln \tan x) \). Since \( \sqrt{x} \ln x \to 0 \), this tends to \( e^{0} = 1 \).
2. (\( \sec \) x)^1/x = \( \exp(\ln \sec x / x) \) and \( \ln \sec x \approx x^{2}/2 \), so the exponent → 0 and the term tends to 1.
3. Sum = 2.

Teacher's Note:
1. Power-times-log limits like \( \sqrt{x} \ln x \) tend to zero.