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

Given: lim x → a x - a x - a By putting directly the limits we get the indeterminate form 0 0 . ∴ lim x → a x − a x − a = lim x → a x - a x - a × x + a x + a (Rationalizing the denominator) = lim x → a x - a x + a x - a ∵ a 2 - b 2 = a - b a + b = lim x → a x + a = a + a = 2 a

Evaluate the limit lim x → 0 2 - x - 2 + x x

We have lim x → 0 2 - x - 2 + x x By putting the limit directly we get the indeterminate form 0 0 ∴ lim x → 0 2 − x − 2 + x x = lim x → 0 2 - x - 2 + x x × 2 - x + 2 + x 2 - x + 2 + x (Rationalizing Numerator) = lim x → 0 2 - x - 2 + x x 2 - x + 2 + x ∵ a - b a + b = a 2 - b 2 = lim x → 0 - 2 x x 2 - x + 2 + x = lim x → 0 - 2 2 - x + 2 + x = - 2 2 + 2 = - 1 2

Evaluate the limit lim h → 0 x + h - x h , x ≠ 0

We have, l i m h → 0 x + h - x h By putting the limit directly we get the indeterminate form 0 0 . ∴ lim h → 0 x + h − x h = l i m h → 0 x + h - x h × x + h + x x + h + x (Rationalizing the numerator) = l i m h → 0 x + h - x h x + h + x ∵ a - b a + b = a 2 - b 2 = l i m h → 0 h h x + h + x = l i m h → 0 1 x + h + x = 1 x + x = 1 2 x

Evaluate the limit lim x → 6 5 + 2 x - 3 + 2 x 2 - 6

We have, l i m x → 6 5 + 2 x - 3 + 2 x 2 - 6 By putting the limit directly we get the indeterminate form 0 0 . ∴ lim x → 6 5 + 2 x − ( 3 + 2 ) x 2 − 6 = l i m x → 6 5 + 2 x - 3 + 2 2 x 2 - 6 = l i m x → 6 5 + 2 x - 5 + 2 6 x 2 - 6 = l i m x → 6 5 + 2 x - 5 + 2 6 x 2 - 6 × 5 + 2 x + 5 + 2 6 5 + 2 x + 5 + 2 6 (Rationalizing the numerator) = l i m x → 6 5 + 2 x - 5 + 2 6 x 2 - 6 5 + 2 x + 5 + 2 6 ∵ a - b a + b = a 2 - b 2 = l i m x → 6 2 x - 6 x - 6 x + 6 5 + 2 x + 5 + 2 6 ∵ ( a − b ) ( a + b ) = a 2 − b 2 = l i m x → 6 2 x + 6 5 + 2 x + 5 + 2 6 = 2 6 + 6 5 + 2 6 + 5 + 2 6 = 2 2 6 · 2 5 + 2 6 = 1 2 6 3 + 2 + 2 6 = 1 2 6 3 + 2 2 = 1 2 6 3 + 2 = 3 - 2 2 6 3 + 2 3 - 2 = 3 - 2 2 6 3 - 2 ∵ ( a − b ) ( a + b ) = a 2 − b 2 = 3 - 2 2 6

Evaluate the limit lim x → 2 3 + 2 x - 2 + 1 x 2 - 2

We have, lim x → 2 3 + 2 x - 2 + 1 x 2 - 2 By putting the limit directly we get the indeterminate form 0 0 ∴ lim x → 2 3 + 2 x − ( 2 + 1 ) x 2 − 2 = lim x → 2 3 + 2 x - 2 + 1 2 x 2 - 2 = lim x → 2 3 + 2 x - 3 + 2 2 x 2 - 2 = lim x → 2 3 + 2 x - 3 + 2 2 x 2 - 2 × 3 + 2 x + 3 + 2 2 3 + 2 x + 3 + 2 2 (Rationalising numerator) = lim x → 2 3 + 2 x - 3 + 2 2 x 2 - 2 3 + 2 x + 3 + 2 2 ∵ a - b a + b = a 2 - b 2 = lim x → 2 2 x - 2 x - 2 x + 2 3 + 2 x + 3 + 2 2 ∵ ( a − b ) ( a + b ) = a 2 − b 2 = lim x → 2 2 x + 2 3 + 2 x + 3 + 2 2 = 2 2 + 2 3 + 2 2 + 3 + 2 2 = 2 2 2 · 2 · 3 + 2 2 = 1 2 2 2 + 1 + 2 2 = 1 2 2 2 + 1 2 = 1 2 2 2 + 1 = 1 2 2 2 + 1 × 2 - 1 2 - 1 = 2 - 1 2 2 2 - 1 = 2 - 1 2 2

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MATHEMATICS for CLASS XI>Chapter 29 - Limits>EXERCISE 29.4

R D Sharma Solutions for Chapter: Limits, Exercise 4: EXERCISE 29.4

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Limits, Exercise 4: EXERCISE 29.4

Attempt the practice questions on Chapter 29: Limits, Exercise 4: EXERCISE 29.4 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.