Evaluate the limit lim x → a x 2 7 - a 2 7 x - a

We have, l i m x → a x 2 7 - a 2 7 x - a By putting the limits directly we get the indeterminate form 0 0 . ∴ lim x → a x 2 7 − a 2 7 x − a = 2 7 a 2 7 - 1 ∵ lim z → p z n - p n z - p = n p n - 1 = 2 7 a - 5 7

Evaluate the limit lim x → a x 5 7 - a 5 7 x 2 7 - a 2 7

We have, l i m x → a x 5 7 - a 5 7 x 2 7 - a 2 7 By putting the limits directly we get the indeterminate form 0 0 . So, lim x → a x 5 7 − a 5 7 x 2 7 − a 2 7 = l i m x → a x 5 7 - a 5 7 x - a x 2 7 - a 2 7 x - a = l i m x → a x 5 7 - a 5 7 x - a ÷ lim x → a ⁡ x 2 7 - a 2 7 x - a = 5 7 a 5 7 - 1 ÷ 2 7 a 2 7 - 1 ∵ lim z → p z n - p n z - p = n p n - 1 = 5 7 a - 2 7 2 7 a - 5 7 = 5 2 a 3 7

Evaluate the limit lim x → a x 2 3 - a 2 3 x 3 4 - a 3 4

We have, lim x → a x 2 3 - a 2 3 x 3 4 - a 3 4 By putting the limits directly we get the indeterminate form 0 0 . So, lim x → a x 2 3 − a 2 3 x 3 4 − a 3 4 = l i m x → a x 2 3 - a 2 3 x - a x 3 4 - a 3 4 x - a = lim x → a x 2 3 − a 2 3 x − a ÷ lim x → a x 3 4 − a 3 4 x − a = 2 3 a 2 3 - 1 ÷ 3 4 a 3 4 - 1 ∵ lim z → p z n - p n z - p = n p n - 1 = 2 3 a - 1 3 3 4 a - 1 4 = 8 9 a - 1 12

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MATHEMATICS for CLASS XI>Chapter 29 - Limits>EXERCISE 29.5

R D Sharma Solutions for Chapter: Limits, Exercise 5: EXERCISE 29.5

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Limits, Exercise 5: EXERCISE 29.5

Attempt the practice questions on Chapter 29: Limits, Exercise 5: EXERCISE 29.5 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.