Class 11 Mathematics Limits and Derivatives MCQs Set 19

Answer: (d) 88

Explanation:
1. The limit of a polynomial is found by direct substitution.
2. 4(27) − 2(9) − 3 + 1 = 108 − 18 − 3 + 1 = 88

Teacher's Note:
1. For polynomials, lim f(x) as x → a is simply f(a).
2. Work out the powers first to avoid sign slips.

Answer: (a) 1

Explanation:
1. For x → 1⁺ (x slightly greater than 1), f(x) = 2 − x².
2. RHL \( = \lim_{x \to 1^+} (2 - x^2) = 2 - 1 = 1 \)

Teacher's Note:
1. For a right hand limit use the piece of the function defined for x > a.
2. Here the left hand limit is also 1 + 1 = 2, so the limit at x = 1 does not exist.

Answer: (b) 7

Explanation:
1. For x → 2⁻, f(x) = 2x + 3.
2. LHL \( = \lim_{x \to 2^-} (2x + 3) = 4 + 3 = 7 \)

Teacher's Note:
1. For a left hand limit use the piece valid for x < a.
2. The right hand limit is 2 + 5 = 7 too, so here the limit exists and equals 7.

Answer: (b) −1

Explanation:
1. For x < 3, x − 3 is negative, so |x − 3| = −(x − 3).
2. Hence \( f(x) = \frac{-(x - 3)}{x - 3} = -1 \) for x < 3, and LHL = −1.

Teacher's Note:
1. Open the modulus with the correct sign on each side of the point before taking the limit.
2. The value f(3) = 0 does not affect the left hand limit.

Answer: (b) \( \sqrt{2} \)

Explanation:
1. Direct substitution does not give 0/0, so substitute x = 0.
2. \( \frac{\sqrt{2} + \sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \)

Teacher's Note:
1. Always try direct substitution first; only a 0/0 form needs further work.
2. Here the denominator is 2, not 0, so no rationalising is needed.

Answer: (a) \( 3 + 2\sqrt{3} \)

Explanation:
1. At x = 1 the denominator is \( 3 - \sqrt{12} = 3 - 2\sqrt{3} \ne 0 \), so substitute directly.
2. \( \frac{-3}{3 - 2\sqrt{3}} = \frac{-3(3 + 2\sqrt{3})}{9 - 12} = 3 + 2\sqrt{3} \)

Teacher's Note:
1. Rationalise the final answer by multiplying by the conjugate 3 + 2√3.
2. √12 = 2√3; simplifying the surd first keeps the arithmetic clean.

Answer: (b) \( \frac{-1}{10} \)

Explanation:
1. At x = 1 we get 0/0. Factorise: 2x² + x − 3 = (x − 1)(2x + 3) and x − 1 = (√x − 1)(√x + 1).
2. \( \frac{(\sqrt{x} - 1)(2x - 3)}{(\sqrt{x} - 1)(\sqrt{x} + 1)(2x + 3)} = \frac{2x - 3}{(\sqrt{x} + 1)(2x + 3)} \)
3. At x = 1: \( \frac{-1}{2 \times 5} = -\frac{1}{10} \)

Teacher's Note:
1. Writing x − 1 as (√x − 1)(√x + 1) lets the square-root factor cancel.
2. Factorise the quadratic by splitting the middle term: 2x² + 3x − 2x − 3.

Answer: (b) 2

Explanation:
1. \( \frac{4x^2 - 1}{2x - 1} = \frac{(2x - 1)(2x + 1)}{2x - 1} = 2x + 1 \) for x ≠ 1/2
2. Limit \( = 2 \times \frac{1}{2} + 1 = 2 \)

Teacher's Note:
1. Recognise 4x² − 1 as a difference of squares.
2. Cancelling (2x − 1) is allowed because x only approaches 1/2.

Answer: (c) 12

Explanation:
1. Use \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) with n = 3, a = 2.
2. Limit = 3 × 2² = 12.

Teacher's Note:
1. The standard limit \( \frac{x^n - a^n}{x - a} \to na^{n-1} \) saves factorising.
2. Alternatively x³ − 8 = (x − 2)(x² + 2x + 4), giving 4 + 4 + 4 = 12.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. \( \frac{2}{1 - x^2} - \frac{1}{1 - x} = \frac{2 - (1 + x)}{1 - x^2} = \frac{1 - x}{(1 - x)(1 + x)} \)
2. \( = \frac{1}{1 + x} \to \frac{1}{2} \)

Teacher's Note:
1. Combine the fractions over a common denominator before taking the limit.
2. Write 1/(x − 1) as −1/(1 − x) so that the denominators match.

Answer: (a) 0

Explanation:
1. The denominator tends to \( \sqrt{1} + 1 = 2 \ne 0 \).
2. Limit \( = \frac{0}{2} = 0 \)

Teacher's Note:
1. Check the form first: 0/2 is not indeterminate.
2. Do not rationalise when direct substitution already works.

Answer: (a) \( \frac{3}{2} \)

Explanation:
1. Divide numerator and denominator by x − 1: \( \frac{\frac{x^{15} - 1}{x - 1}}{\frac{x^{10} - 1}{x - 1}} \)
2. Using \( \frac{x^n - 1}{x - 1} \to n \): limit \( = \frac{15}{10} = \frac{3}{2} \)

Teacher's Note:
1. Dividing both parts by (x − a) turns the expression into a ratio of two standard limits.
2. With a = 1, na^{n−1} is just n.

Answer: (c) 5

Explanation:
1. \( \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = n \cdot 2^{n-1} \)
2. \( n \cdot 2^{n-1} = 80 = 5 \times 2^4 \), so n = 5.

Teacher's Note:
1. Write 80 as n × 2ⁿ⁻¹ by trying the options or by factorising 80.
2. Check: 5 × 2⁴ = 80.

Answer: (b) \( \frac{1}{3(2)^{2/3}} \)

Explanation:
1. Put y = x + 2; as x → 0, y → 2: \( \lim_{y \to 2} \frac{y^{1/3} - 2^{1/3}}{y - 2} \)
2. \( = \frac{1}{3} \cdot 2^{\frac{1}{3} - 1} = \frac{1}{3 \cdot 2^{2/3}} \)

Teacher's Note:
1. The result na^{n−1} holds for rational n as well, here n = 1/3.
2. Changing the variable to y = x + 2 makes the standard form visible.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. Rationalise: \( \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \frac{x}{x(\sqrt{1 + x} + 1)} \)
2. \( = \frac{1}{\sqrt{1 + x} + 1} \to \frac{1}{2} \)

Teacher's Note:
1. Multiply by the conjugate when a 0/0 form contains a square root.
2. The same answer follows from the standard limit with n = 1/2 and a = 1.

Answer: (b) \( \frac{7}{5} \)

Explanation:
1. \( \frac{\sin 7x}{\tan 5x} = \frac{\sin 7x}{7x} \cdot \frac{5x}{\tan 5x} \cdot \frac{7}{5} \)
2. Each fraction tends to 1, so the limit is \( \frac{7}{5} \).

Teacher's Note:
1. Multiply and divide by the angle inside each trig function to create standard limits.
2. For small x, sin kx and tan kx both behave like kx.

Answer: (b) 2

Explanation:
1. \( \frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \)
2. Limit = 1 × 1 × 2 = 2.

Teacher's Note:
1. In general \( \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b} \).
2. Alternatively sin 4x = 2 sin 2x cos 2x, giving 2 cos 2x → 2.

Answer: (a) 1

Explanation:
1. x° = \( \frac{\pi x}{180} \) radians; let \( t = \frac{\pi x}{180} \to 0 \).
2. \( \frac{\tan x^\circ}{x^\circ} = \frac{\tan t}{t} \to 1 \)

Teacher's Note:
1. The standard limits hold when the angle and the denominator are the same quantity.
2. By contrast, \( \lim_{x \to 0} \frac{\sin x^\circ}{x} = \frac{\pi}{180} \) because the denominator is not in the same unit.

Answer: (a) 0

Explanation:
1. \( 1 - \cos 4x = 2\sin^2 2x \), so \( \frac{1 - \cos 4x}{x} = 2 \cdot \frac{\sin 2x}{2x} \cdot 2 \sin 2x \)
2. As x → 0 this tends to 2 × 1 × 2 × 0 = 0.

Teacher's Note:
1. Remember \( \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \), while \( \frac{1 - \cos x}{x^2} \to \frac{1}{2} \).
2. The power of x in the denominator decides which result to use.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. \( \frac{\sin x}{x} \cdot \frac{1}{1 + \cos x} \to 1 \times \frac{1}{1 + 1} = \frac{1}{2} \)

Teacher's Note:
1. Split the expression into a standard limit times a part that can be substituted directly.
2. cos 0 = 1, so the second factor tends to 1/2.

Answer: (d) 2, 2

Explanation:
1. \( \sin(2 + x) - \sin(2 - x) = 2\cos 2 \sin x \)
2. \( \frac{2\cos 2 \sin x}{x} \to 2\cos 2 \)
3. So p = 2 and q = 2.

Teacher's Note:
1. Use sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2).
2. After the identity, only the standard limit sin x / x remains.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. \( \tan x - \sin x = \frac{\sin x(1 - \cos x)}{\cos x} \)
2. \( \frac{\sin x(1 - \cos x)}{\cos x \sin^3 x} = \frac{1 - \cos x}{\cos x(1 - \cos^2 x)} = \frac{1}{\cos x(1 + \cos x)} \)
3. As x → 0 this tends to \( \frac{1}{1 \times 2} = \frac{1}{2} \).

Teacher's Note:
1. Write tan x as sin x / cos x and take sin x common.
2. sin² x = (1 − cos x)(1 + cos x) is the key factorisation.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. Divide numerator and denominator by x: \( \frac{\frac{\tan 2x}{x} - 1}{3 - \frac{\sin x}{x}} \)
2. \( \to \frac{2 - 1}{3 - 1} = \frac{1}{2} \)

Teacher's Note:
1. \( \frac{\tan 2x}{x} = 2 \cdot \frac{\tan 2x}{2x} \to 2 \).
2. Dividing every term by x is the standard way to handle sums in such limits.

Answer: (d) 2

Explanation:
1. Put \( x = \frac{\pi}{2} + h \), h → 0: tan 2x = tan(π + 2h) = tan 2h.
2. \( \lim_{h \to 0} \frac{\tan 2h}{h} = 2 \)

Teacher's Note:
1. When x → a with a ≠ 0, substitute x = a + h to bring the limit to h → 0.
2. tan has period π, so tan(π + 2h) = tan 2h.

Answer: (d) 2

Explanation:
1. \( \sec^2 x - 2 = 1 + \tan^2 x - 2 = \tan^2 x - 1 \)
2. \( \frac{(\tan x - 1)(\tan x + 1)}{\tan x - 1} = \tan x + 1 \to 1 + 1 = 2 \)

Teacher's Note:
1. Use sec² x = 1 + tan² x to express everything in tan x.
2. Then it is an ordinary factorisation of a difference of squares.