CBSE Class 11 Mathematics Limits and Derivatives Competitive Exam MCQs Set 01

Answer: (a) 3

Explanation:
1. At x = 2 both numerator and denominator are 0, so factorise.
2. \( \frac{x^3 - 8}{x^2 - 4} = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} = \frac{x^2 + 2x + 4}{x + 2} \)
3. Put x = 2: \( \frac{4 + 4 + 4}{2 + 2} = 3 \)

Teacher's Note:
1. When direct substitution gives 0/0, the factor (x − a) is common to numerator and denominator and can be cancelled.
2. The standard result \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) gives the same answer: (3 × 2²)/(2 × 2) = 3.

Answer: (d) \( 1 + \frac{1}{\sqrt{2}} \)

Explanation:
1. Split the fraction: \( 1 + \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} \)
2. \( \frac{\sqrt{x - 1}}{\sqrt{(x - 1)(x + 1)}} = \frac{1}{\sqrt{x + 1}} \) for x > 1
3. As x → 1⁺ the expression tends to \( 1 + \frac{1}{\sqrt{2}} \)

Teacher's Note:
1. Splitting the fraction into two parts is often quicker than rationalising.
2. The limit is taken as x → 1⁺ because √(x − 1) is real only for x ≥ 1.

Answer: (b) \( \frac{2}{5} \)

Explanation:
1. x² − 3 = (x − √3)(x + √3)
2. x² + 3√3x − 12 = (x − √3)(x + 4√3)
3. Cancel (x − √3) and put x = √3: \( \frac{2\sqrt{3}}{5\sqrt{3}} = \frac{2}{5} \)

Teacher's Note:
1. To factorise x² + 3√3x − 12, look for two numbers with sum 3√3 and product −12; they are 4√3 and −√3.
2. Irrational roots are handled exactly like integer roots in the factorisation method.

Answer: (a) 2

Explanation:
1. Let y = x^−1/3; as x → 1, y → 1.
2. \( \frac{1 - y^2}{1 - y} = \frac{(1 - y)(1 + y)}{1 - y} = 1 + y \)
3. Limit = 1 + 1 = 2

Teacher's Note:
1. A substitution like y = x^(−1/3) turns fractional powers into an ordinary polynomial ratio.
2. As x → 1, every power of x also tends to 1, so the new variable tends to 1.

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

Explanation:
1. 2x² + x − 3 = (x − 1)(2x + 3) and x − 1 = (√x − 1)(√x + 1)
2. Expression = \( \frac{2x - 3}{(\sqrt{x} + 1)(2x + 3)} \)
3. Put x = 1: \( \frac{-1}{2 \times 5} = -\frac{1}{10} \)

Teacher's Note:
1. Writing x − 1 as (√x − 1)(√x + 1) is a quick way to cancel a factor containing √x.
2. Always factorise the quadratic in the denominator first to see which factor vanishes.

Answer: (d) \( -\frac{1}{56} \)

Explanation:
1. Multiply numerator and denominator by 2 + √(x − 3): numerator becomes 4 − (x − 3) = 7 − x.
2. Expression = \( \frac{-(x - 7)}{(x - 7)(x + 7)(2 + \sqrt{x - 3})} = \frac{-1}{(x + 7)(2 + \sqrt{x - 3})} \)
3. Put x = 7: \( \frac{-1}{14 \times 4} = -\frac{1}{56} \)

Teacher's Note:
1. Rationalise the part containing the square root so that the troublesome factor (x − 7) appears.
2. Watch the sign: 7 − x = −(x − 7), which makes the answer negative.

Answer: (a) 3

Explanation:
1. Rationalise numerator: \( \sqrt{x^2 + 1} - 1 = \frac{x^2}{\sqrt{x^2 + 1} + 1} \)
2. Rationalise denominator: \( \sqrt{x^2 + 9} - 3 = \frac{x^2}{\sqrt{x^2 + 9} + 3} \)
3. Expression = \( \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 1} + 1} \to \frac{6}{2} = 3 \)

Teacher's Note:
1. When both numerator and denominator contain roots, rationalise both of them.
2. After rationalising, the common factor x² cancels and direct substitution works.

Answer: (b) \( \frac{1}{4b\sqrt{b - a}} \)

Explanation:
1. Multiply by \( \sqrt{x - a} + \sqrt{b - a} \): numerator becomes (x − a) − (b − a) = x − b.
2. Expression = \( \frac{1}{(x + b)(\sqrt{x - a} + \sqrt{b - a})} \)
3. Put x = b: \( \frac{1}{2b \cdot 2\sqrt{b - a}} = \frac{1}{4b\sqrt{b - a}} \)

Teacher's Note:
1. For √(b − a) to be real we need b > a, which already rules out the options containing √(a − b).
2. x² − b² = (x − b)(x + b) supplies the factor that cancels after rationalising.

Answer: (a) \( \frac{1}{8\sqrt{3}} \)

Explanation:
1. Multiply by \( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \): numerator becomes \( \sqrt{2 + x} - 2 \).
2. Multiply by \( \sqrt{2 + x} + 2 \): this becomes x − 2, which cancels with the denominator.
3. Expression = \( \frac{1}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(\sqrt{2 + x} + 2)} \)
4. Put x = 2: \( \frac{1}{2\sqrt{3} \times 4} = \frac{1}{8\sqrt{3}} \)

Teacher's Note:
1. For nested square roots, rationalise one layer at a time, starting from the outer root.
2. Each rationalisation leaves a factor like (√(2 + x) + 2) that simply becomes a number at the end.

Answer: (d) 0

Explanation:
1. Highest power in the denominator is x³; divide numerator and denominator by x³.
2. Expression = \( \frac{\left(1 + \frac{1}{x}\right)\left(3 + \frac{4}{x}\right) \cdot \frac{1}{x}}{1 - \frac{8}{x}} \)
3. As x → ∞: \( \frac{1 \times 3 \times 0}{1} = 0 \)

Teacher's Note:
1. If the degree of the numerator is less than the degree of the denominator, the limit at infinity is 0.
2. Here the numerator has degree 2 and the denominator degree 3.

Answer: (a) −1

Explanation:
1. Both numerator and denominator have degree 45; divide each by x⁴⁵.
2. Expression = \( \frac{\left(\frac{2}{x} + 1\right)^{40}\left(\frac{4}{x} + 1\right)^5}{\left(\frac{2}{x} - 1\right)^{45}} \)
3. As x → ∞: \( \frac{1 \cdot 1}{(-1)^{45}} = -1 \)

Teacher's Note:
1. For equal degrees the limit is the ratio of the leading coefficients.
2. The leading coefficient of (2 − x)⁴⁵ is (−1)⁴⁵ = −1, which gives the negative answer.

Answer: (a) 0

Explanation:
1. (n + 2)! + (n + 1)! = (n + 1)!(n + 2 + 1) = (n + 1)!(n + 3)
2. (n + 3)! = (n + 3)(n + 2)(n + 1)!
3. Expression = \( \frac{1}{n + 2} \to 0 \) as n → ∞

Teacher's Note:
1. Take the smallest factorial, (n + 1)!, common from every term.
2. After cancelling, the expression is 1/(n + 2), which becomes smaller and smaller as n grows.

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

Explanation:
1. 1 + 2 + … + n = \( \frac{n(n + 1)}{2} \) (sum of an AP)
2. Expression = \( \frac{n^2 + n}{2(n^2 + 100)} = \frac{1 + \frac{1}{n}}{2\left(1 + \frac{100}{n^2}\right)} \)
3. As n → ∞ this tends to \( \frac{1}{2} \)

Teacher's Note:
1. Replace the sum by its closed form before taking the limit.
2. The large constant 100 does not matter: only the highest powers decide a limit at infinity.

Answer: (c) 1

Explanation:
1. This is a GP with first term 1/2 and common ratio 1/2.
2. Sum of n terms = \( \frac{\frac{1}{2}\left(1 - \frac{1}{2^n}\right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} \)
3. As n → ∞, \( \frac{1}{2^n} \to 0 \), so the limit is 1.

Teacher's Note:
1. The limit of the partial sums of a GP with |r| < 1 is the sum of the infinite GP, a/(1 − r).
2. Here a/(1 − r) = (1/2)/(1/2) = 1.

Answer: (d) zero

Explanation:
1. Write 2²ⁿ = 4ⁿ and 3²ⁿ⁺³ = 27 · 9ⁿ; the fastest-growing term is 9ⁿ.
2. Divide by 9ⁿ: \( \frac{5\left(\frac{5}{9}\right)^n + \left(\frac{1}{3}\right)^n - \left(\frac{4}{9}\right)^n}{\left(\frac{5}{9}\right)^n + \left(\frac{2}{9}\right)^n + 27} \)
3. Each \( r^n \to 0 \) since |r| < 1, so the limit is \( \frac{0}{27} = 0 \).

Teacher's Note:
1. Divide numerator and denominator by the fastest-growing term, just as we divide by the highest power of x.
2. If |r| < 1 then rⁿ → 0 as n → ∞, the same fact that makes an infinite GP converge.

Answer: (b) \( \frac{1}{6} \)

Explanation:
1. Multiply and divide by \( 3x - \sqrt{9x^2 - x} \): expression = \( \frac{9x^2 - (9x^2 - x)}{3x - \sqrt{9x^2 - x}} = \frac{x}{3x - \sqrt{9x^2 - x}} \)
2. For x < 0, \( \sqrt{9x^2 - x} = |x|\sqrt{9 - \frac{1}{x}} = -x\sqrt{9 - \frac{1}{x}} \)
3. Expression = \( \frac{x}{3x + x\sqrt{9 - \frac{1}{x}}} = \frac{1}{3 + \sqrt{9 - \frac{1}{x}}} \to \frac{1}{6} \)

Teacher's Note:
1. For x → −∞ remember that √(x²) = |x| = −x; ignoring this sign gives a wrong answer.
2. The form ∞ − ∞ is removed by multiplying with the conjugate expression.

Answer: (c) 18

Explanation:
1. \( \frac{2\sin^2 3x}{x^2} = 2 \times 9 \times \left( \frac{\sin 3x}{3x} \right)^2 \)
2. Since \( \frac{\sin 3x}{3x} \to 1 \), the limit is 2 × 9 × 1 = 18.

Teacher's Note:
1. Make the angle and the denominator match: sin 3x needs 3x below it.
2. The square of x supplies the factor 3² = 9.

Answer: (c) \( \frac{2}{\pi} \)

Explanation:
1. Put x = 1 − h, h → 0: \( \tan\left( \frac{\pi}{2} - \frac{\pi h}{2} \right) = \cot\frac{\pi h}{2} \)
2. Expression = \( \frac{h}{\tan\frac{\pi h}{2}} = \frac{2}{\pi} \cdot \frac{\frac{\pi h}{2}}{\tan\frac{\pi h}{2}} \)
3. Limit = \( \frac{2}{\pi} \times 1 = \frac{2}{\pi} \)

Teacher's Note:
1. Shift the point to 0 by a substitution so that the standard limit tan θ/θ → 1 can be used.
2. tan(π/2 − θ) = cot θ = 1/tan θ.

Answer: (c) 1

Explanation:
1. Divide numerator and denominator by x: \( \sqrt{\frac{1 - \frac{\sin x}{x}}{1 + \frac{\cos^2 x}{x}}} \)
2. sin x and cos² x stay between −1 and 1, so \( \frac{\sin x}{x} \to 0 \) and \( \frac{\cos^2 x}{x} \to 0 \) as x → ∞.
3. Limit = √1 = 1

Teacher's Note:
1. A bounded quantity divided by a quantity that grows without limit tends to 0.
2. Do not use sin x/x → 1 here; that result is for x → 0, not x → ∞.

Answer: (a) 0

Explanation:
1. \( \sec\theta - \tan\theta = \frac{1 - \sin\theta}{\cos\theta} \)
2. Multiply by (1 + sin θ): \( \frac{\cos^2\theta}{\cos\theta(1 + \sin\theta)} = \frac{\cos\theta}{1 + \sin\theta} \)
3. At θ = π/2: \( \frac{0}{2} = 0 \)

Teacher's Note:
1. The form ∞ − ∞ is first combined into a single fraction.
2. Multiplying by 1 + sin θ uses 1 − sin²θ = cos²θ to remove the 0/0 form.

Answer: (a) \( \frac{\pi}{4} \)

Explanation:
1. Put t = \( \frac{\pi}{4x} \); as x → ∞, t → 0 and x = \( \frac{\pi}{4t} \).
2. Expression = \( \frac{\pi}{4} \cos t \cdot \frac{\sin t}{t} \)
3. Limit = \( \frac{\pi}{4} \times 1 \times 1 = \frac{\pi}{4} \)

Teacher's Note:
1. A limit at infinity becomes a limit at 0 with the substitution t = 1/x (here t = π/(4x)).
2. After the substitution the standard result sin t/t → 1 applies directly.

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

Explanation:
1. \( \cot^2 \pi x = \frac{\cos^2 \pi x}{\sin^2 \pi x} = \frac{\cos^2 \pi x}{(1 - \cos \pi x)(1 + \cos \pi x)} \)
2. Expression = \( \frac{\cos^2 \pi x}{1 - \cos \pi x} \)
3. At x = 1, cos π = −1: \( \frac{1}{1 + 1} = \frac{1}{2} \)

Teacher's Note:
1. Writing sin² as (1 − cos)(1 + cos) cancels the factor that vanishes at x = 1.
2. Once the cancelling is done, direct substitution gives the answer.

Answer: (d) \( \frac{4}{3} \)

Explanation:
1. √3 sin θ − cos θ = 2 sin(θ − π/6); with θ = π/6 + h the bracket equals 2 sin h, so the numerator is 4 sin h.
2. Expression = \( \frac{4}{\sqrt{3}} \cdot \frac{\sin h}{h} \cdot \frac{1}{\sqrt{3}\cos h - \sin h} \)
3. Limit = \( \frac{4}{\sqrt{3}} \times 1 \times \frac{1}{\sqrt{3}} = \frac{4}{3} \)

Teacher's Note:
1. An expression a sin θ − b cos θ can be written as one sine using the compound-angle formula.
2. After simplifying, only the standard limit sin h/h → 1 is needed.

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

Explanation:
1. tan θ − sin θ = tan θ(1 − cos θ) = 2 tan θ sin²(θ/2), and (1 − cos 2θ)² = 4 sin⁴θ.
2. Expression = \( \frac{8\theta\tan\theta\sin^2\frac{\theta}{2}}{4\sin^4\theta} = \frac{1}{2} \cdot \frac{\tan\theta}{\theta} \cdot \left( \frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}} \right)^2 \cdot \left( \frac{\theta}{\sin\theta} \right)^4 \)
3. Each bracket tends to 1, so the limit is \( \frac{1}{2} \).

Teacher's Note:
1. Convert 1 − cos into 2 sin² of the half angle; this is the key step in most such limits.
2. Then arrange every sine and tangent next to its own angle so each ratio tends to 1.

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

Explanation:
1. \( \operatorname{cosec} x - \cot x = \frac{1 - \cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2} \)
2. Expression = \( \frac{1}{2} \cdot \frac{\tan\frac{x}{2}}{\frac{x}{2}} \)
3. Limit = \( \frac{1}{2} \)

Teacher's Note:
1. cosec x − cot x simplifies neatly to tan(x/2) using half-angle formulas.
2. Then the standard limit tan θ/θ → 1 finishes the problem.

Answer: (b) 1

Explanation:
1. Rationalise numerator and denominator: \( \frac{(9 - f(x))(3 + \sqrt{x})}{(9 - x)(3 + \sqrt{f(x)})} \)
2. \( \frac{9 - f(x)}{9 - x} = \frac{f(x) - f(9)}{x - 9} \to f'(9) = 1 \), and f(x) → f(9) = 9.
3. Limit = \( 1 \times \frac{3 + 3}{3 + 3} = 1 \)

Teacher's Note:
1. Look for the pattern (f(x) − f(a))/(x − a), which is the first-principle definition of f′(a).
2. A function that is differentiable at a point is also continuous there, so f(x) → f(9).

Answer: (d) −cos a

Explanation:
1. Group: [sin(a + 3h) − sin a] − 3[sin(a + 2h) − sin(a + h)] = 2cos(a + 3h/2) sin(3h/2) − 6cos(a + 3h/2) sin(h/2)
2. = 2cos(a + 3h/2)[sin(3h/2) − 3 sin(h/2)] = −8 cos(a + 3h/2) sin³(h/2), using sin 3θ = 3 sin θ − 4 sin³θ.
3. Divide by h³: \( -\cos\left(a + \frac{3h}{2}\right) \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^3 \to -\cos a \)

Teacher's Note:
1. Pairing the terms so that sin C − sin D can be used produces a common factor cos(a + 3h/2).
2. The triple-angle formula sin 3θ = 3 sin θ − 4 sin³θ then supplies the h³ that cancels.

Answer: (c) \( \frac{4a}{\pi} \)

Explanation:
1. Let \( t = \sqrt{\frac{a - x}{a + x}} \); as x → a (from the left, where the root is real), t → 0.
2. \( \sqrt{a^2 - x^2} = (a + x)t \), so the expression = \( (a + x) \cdot \frac{t}{\tan\frac{\pi t}{2}} = (a + x) \cdot \frac{2}{\pi} \cdot \frac{\frac{\pi t}{2}}{\tan\frac{\pi t}{2}} \)
3. Limit = \( 2a \cdot \frac{2}{\pi} = \frac{4a}{\pi} \)

Teacher's Note:
1. Choose the substitution so that the angle inside cot tends to 0.
2. Since cot θ = 1/tan θ, the standard result tan θ/θ → 1 can be used.

Answer: (c) 2 sin 2

Explanation:
1. Near x = −1, |x| = −x, so x² − |x| = x² + x = x(x + 1).
2. cos 2 − cos 2x = 2 sin(1 + x) sin(x − 1)
3. Expression = \( \frac{\sin(1 + x)}{1 + x} \cdot \frac{2\sin(x - 1)}{x} \to 1 \cdot \frac{2\sin(-2)}{-1} = 2\sin 2 \)

Teacher's Note:
1. Remove the modulus first by checking the sign of x near the point.
2. cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2) produces the factor sin(1 + x) needed for the standard limit.

Answer: (a) 2a sin a + a² cos a

Explanation:
1. Numerator = a²[sin(a + h) − sin a] + (2ah + h²) sin(a + h)
2. \( \frac{a^2[\sin(a + h) - \sin a]}{h} = a^2 \cos\left(a + \frac{h}{2}\right) \cdot \frac{\sin\frac{h}{2}}{\frac{h}{2}} \to a^2 \cos a \)
3. \( \frac{(2ah + h^2)\sin(a + h)}{h} = (2a + h)\sin(a + h) \to 2a \sin a \)
4. Limit = 2a sin a + a² cos a

Teacher's Note:
1. This limit is the derivative of x² sin x at x = a by first principle.
2. The product rule gives the same result: (x²)′ sin x + x² (sin x)′ = 2x sin x + x² cos x.

Answer: (b) −2, −1

Explanation:
1. sin 2x + a sin x = sin x(2cos x + a), so the expression = \( \frac{\sin x}{x} \cdot \frac{2\cos x + a}{x^2} \)
2. For a finite limit the numerator 2cos x + a must vanish at x = 0: 2 + a = 0, a = −2.
3. Then \( \frac{2\cos x - 2}{x^2} = -2 \cdot \frac{1 - \cos x}{x^2} \to -2 \times \frac{1}{2} = -1 \)

Teacher's Note:
1. If the denominator tends to 0 and the limit is finite, the numerator must also tend to 0.
2. (1 − cos x)/x² = 2 sin²(x/2)/x² → 1/2 is a useful result to remember.

Answer: (a) a rational number; (b) 2

Explanation:
1. The terms are \( 2^{1/2}, 2^{1/2 + 1/4}, 2^{1/2 + 1/4 + 1/8}, \ldots \); the n^th term is \( 2^{\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}} \).
2. The exponent is a GP sum: \( 1 - \frac{1}{2^n} \to 1 \).
3. So the limit is 2¹ = 2, which is a rational number; options (a) and (b) are correct.

Teacher's Note:
1. Write nested roots as powers; each new root halves the previous exponent.
2. The exponents form a GP with first term 1/2 and ratio 1/2, whose infinite sum is 1.