Evaluate the following limit: lim x → 0 ( a + x ) 2 sin ( a + x ) - a 2 sin a x .

Let L = lim x → 0 ( a + x ) 2 sin ( a + x ) - a 2 sin a x Therefore, L = lim x → 0 ( a 2 + 2 a x + x 2 ) sin ( a + x ) - a 2 sin a x ⇒ L = lim x → 0 a 2 sin ( a + x ) + 2 a x sin ( a + x ) + x 2 sin ( a + x ) - a 2 sin a x ⇒ L = lim x → 0 a 2 sin ( a + x ) - sin a + 2 a x sin ( a + x ) + x 2 sin ( a + x ) x ⇒ L = lim x → 0 a 2 sin ( a + x ) - sin a x + lim x → 0 2 a sin a + x + lim x → 0 x sin a + x {Since x → 0 , therefore, x ≠ 0 ) undefined ⇒ L = 2 a 2 lim x → 0 cos 2 a + x 2 sin x 2 x + lim x → 0 2 a sin a + x + lim x → 0 x sin a + x ⇒ L = 2 a 2 lim x → 0 cos 2 a + x 2 × lim x → 0 sin x 2 x 2 × 2 + lim x → 0 2 a sin a + x + lim x → 0 x sin a + x ⇒ L = a 2 lim x → 0 cos 2 a + x 2 × lim x 2 → 0 sin x 2 x 2 + lim x → 0 2 a sin a + x + lim x → 0 x sin a + x ⇒ L = a 2 cos a × 1 + 2 a sin a + 0 = 2 a sin a + a 2 cos a ∵ lim θ → 0 sin θ θ = 1 Hence, lim x → 0 ( a + x ) 2 sin ( a + x ) - a 2 sin a x = 2 a sin a + a 2 cos a .

Evaluate the following limit: lim x → 0 e 3 + x - sin x - e 3 x .

Let L = lim x → 0 e 3 + x - sin x - e 3 x Therefore, L = lim x → 0 e 3 + x - e 3 - sin x x ⇒ L = - lim x → 0 sin x x + lim x → 0 e 3 + x - e 3 x {Since x → 0 , therefore, x ≠ 0 ) ⇒ L = - lim x → 0 sin x x + e 3 lim x → 0 e x - 1 x ⇒ L = - 1 + e 3 ∵ lim θ → 0 sin θ θ = 1 , lim x → 0 e x - 1 x = 1 Hence, lim x → 0 e 3 + x - sin x - e 3 x = e 3 - 1 .

E M B I B E

WBCHSE Mathematics for Class 11>Calculus>Chapter 23 - Limits>EXERCISE 1B

R S Aggarwal Solutions for Exercise 2: EXERCISE 1B

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE 1B

Attempt the practice questions from Exercise 2: EXERCISE 1B with hints and solutions to strengthen your understanding. WBCHSE Mathematics for Class 11 solutions are prepared by Experienced Embibe Experts.