Question

Lim x tends to 2 (x^2-x-
2)^20/(x^3-12x+16)^10 ?

Answer 1

Solution :

Here, x² - x - 2

= x² - 2x + x - 2

= x (x - 2) + 1 (x - 2)

= (x - 2) (x + 1)

and x³ - 12x + 16

= x³ - 2x² + 2x² - 4x - 8x + 16

= x² (x - 2) + 2x (x - 2) - 8 (x - 2)

= (x - 2) (x² + 2x - 8)

= (x - 2) (x² + 4x - 2x - 8)

= (x - 2) {x (x + 4) - 2 (x + 4) }

= (x - 2) (x - 2) (x + 4)

= (x - 2)² (x + 4)

∴ (x² - x - 2)²⁰/(x³ - 12x + 16)¹⁰

= {(x - 2) (x + 1)}²⁰/{(x - 2)² (x + 4)}¹⁰

= {(x - 2)²⁰ (x + 1)²⁰}/{(x - 2)²⁰ (x + 4)¹⁰}

= (x + 1)²⁰/(x + 4)¹⁰ ,

since x ⇢ 2 does not imply x = 2

∴ lim (x ⇢ 2) (x² - x - 2)²⁰/(x³ - 12x + 16)¹⁰

= lim (x ⇢ 2) (x + 1)²⁰/(x + 4)¹⁰

= (2 + 1)²⁰/(2 + 4)¹⁰

= 3²⁰/6¹⁰

= 3²⁰/(3 * 2)¹⁰

= 3²⁰/(3¹⁰ * 2¹⁰)

= (3²⁰ ⁻ ¹⁰)/2¹⁰

= 3¹⁰/2¹⁰

= (3/2)¹⁰ ( Ans. )