MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers

Answer: (d) e^-1/2
Hint:
Given, Lim_x→0 (cos x)^{cot² x}
= Lim_x→0 (1 + cos x – 1)^{cot² x}
= eLim_x→0 (cos x – 1) × cot² x
= eLim_x→0 (cos x – 1)/tan² x
= e^-1/2

Answer: (c) 2 cos a
Hint:
Given, Lim_x→0 {sin (a + x) – sin (a – x)}/x
= Lim_x→0 {2 × cos a × sin x}/x
= 2 × cos a × Lim_x→0 sin x/x
= 2 cos a

Answer: (b) 1
Hint:
Given, Lim_x→-1 [1 + x + x² + ……….+ x¹⁰]
= 1 + (-1) + (-1)² + ……….+ (-1)¹⁰
= 1 – 1 + 1 – ……. + 1
= 1

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2× tan^-1 x)/x
= 2 × 1
= 2

Answer: (a) 0
Hint:
We know that
log(1 – x) = -x – x²/2 – x³/3 – ……..
Now,
Lim_x→0 log(1 – x) = Lim_x→0 {-x – x²/2 – x³/3 – ……..}
⇒ Lim_x→0 log(1 – x) = Lim_x→0 {-x} – Lim_x→0 {x²/2} – Lim_x→0 {x³/3} – ……..
⇒ Lim_x→0 log(1 – x) = 0

Answer: (c) log (a/b)
Hint:
Given, Lim_x→0 {(a^x – b^x)/ x}
= Lim_x→0 {(a^x – b^x – 1 + 1)/ x}
= Lim_x→0 {(a^x – 1) – (b^x – 1)}/ x
= Lim_x→0 {(a^x – 1)/x – (b^x – 1)/x}
= Lim_x→0 (a^x – 1)/x – Lim_x→0 (b^x – 1)/x
= log a – log b
= log (a/b)

Answer: (d) x × tan x × sec x + sec x
Hint:
Given, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y
= lim_y→0 {x sec (x + y) + y sec (x + y) – x×sec x}/y
= lim_y→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= lim_y→0 x{ sec (x + y) – sec x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 x{1/cos (x + y) – 1/cos x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{cos x – cos (x + y)} × x/{y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 {sin (x + y/2) × lim_y→0 {sin(y/2)/(2y/2)} × lim_y→0 { x/{y×cos (x + y)×cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y = x × tan x × sec x + sec x

Answer: (c) e⁵
Hint:
Given, Lim_y→∞ {(x + 6)/(x + 1)}(x + 4)
= Lim_y→∞ {1 + 5/(x + 1)}(x + 4)
= e^{Lim_y→∞ 5(x + 4)/(x + 1)}
= e^{Lim_y→∞ 5(1 + 4/x)/(1 + 1/x)}
= e^{5(1 + 4/∞)/(1 + 1/∞)}
= e^{5/(1 + 0)}
= e⁵

Answer: (d) A
Hint:
Let y = [1+(1/x)] /[1-(1/x)]
then dy/dx = [{1-(1/x)}*(-1/x²)]/[{1+(1/x)}*(1/x²)]
= (1/x²) [(1/x) -1 – 1 – (1/x)]/[1-(1/x)]²
= [-2/x²]/[(x-1)/x]²
= -2/(x-1)²

Answer: (d) -x – x²/2 – x³/3 – ……..
Hint:
log(1 – x) = -x – x²/2 – x³/3 – ……..

Answer: (b) 0
Hint:
Given, f(x) = x × sin(1/x)
Now, Lim_x→0 f(x) = Lim_x→0 x × sin(1/x)
⇒ Lim_x→0 f(x) = 0

Answer: (a) 0
Hint:
Given, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³
= Lim_n→∞ [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Lim_n→∞ [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Lim_n→∞ [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n⁴ × {(1 + 1/n)/2}²]
⇒ Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/∞)×(2 + 1/∞)}/6]/[∞×{(1 + 1/∞)/2}²
= [{(1 + 0)×(2 + 0)}/6]/∞ {since 1/∞ = 0}
= {(1 × 2)/6}/∞
= (2/6)/∞
= (1/3)/∞
= 0
So, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³ = 0

Answer: (a) 0
Hint:
Lim_n→∞ (sin x/x) = Lim_y→0 {y × sin (1/y)} = 0

Answer: (b) 1
Hint:
We know that
a^x = 1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + ………..
Now,
Lim_x→0 a^x = Lim_x→0 {1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}
⇒ Lim_x→0 a^x = Lim_x→0 1 + Lim_x→0 {x/1! × (log a)} + Lim_x→0 {x² /2! × (log a)²}+ ………
⇒ Lim_x→0 a^x = 1

Answer: (b) 1
Hint:
Given, Lim_x→0 f(x) exists
⇒ Lim_x→0 – f(x) = Lim_x→0 + f(x)
⇒ Lim_x→0 (x + k) = Lim_x→0 cos x
⇒ k = cos 0
⇒ k = 1

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2 × tan^-1 x)/x
= 2 × 1
= 2

Answer: (c) a/b
Hint:
Given, Lim_x→0 sin (ax)/bx
= Lim_x→0 [{sin (ax)/ax} × (ax/bx)]
⇒ (a/b) Lim_x→0 sin (ax)/ax
= a/b

Answer: (c) a
Hint:
Given, Lim_x→0 {log(1 + ax)}/x
= Lim_x→0 {ax – (ax)² /2 + (ax)³ /3 – (ax)⁴ /4 + …….}/x
= Lim_x→0 {ax – a² x² /2 + a³ x³ /3 – a⁴ x⁴ /4 + …….}/x
= Lim_x→0 {a – a² x /2 + a³ x² /3 – a⁴ x³ /4 + …….}
= a – 0
= a

Answer: (c) -1/x²
Hint:
Given, f(x) = (x + 1)/x
Now, df(x)/dx = d{(x + 1)/x}/dx
= {1 × x – (x + 1)×1}/x²
= (x – x – 1)/x²
= -1/x²

Answer: (d) 3/2
Hint:
Given, Lim_x→0 (e^x² – cos x)/x²
= Lim_x→0 (e^x² – cos x -1 + 1)/x²
= Lim_x→0 {(e^x² – 1)/x² + (1 – cos x)}/x²
= Lim_x→0 {(e^x² – 1)/x² + Lim_x→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2