EVALUATING LIMITS BY FACTORING Finding a limit by factoring is a technique to find limit by cancelling out common factors.
If we apply limit without factoring and cancelling the common terms, we may get indeterminant form as answers. So, doing possible simplification is important.
Evaluate the following limits. Show all work.
Problem 1 :
x ⇾ 6 x 2 - 36 x - 6 Solution :
Given , x ⇾ 6 x 2 - 36 x - 6 = x ⇾ 6 x 2 - 6 2 x - 6 = x ⇾ 6 ( x + 6 ) ( x - 6 ) x - 6 = x ⇾ 6 x + 6 = 6 + 6 = 12 Problem 2 :
x ⇾ 3 9 - x 2 3 - x Solution :
Given , x ⇾ 3 9 - x 2 3 - x = x ⇾ 3 3 2 - x 2 3 - x = x ⇾ 3 ( 3 + x ) ( 3 - x ) 3 - x = x ⇾ 3 3 + x = 3 + 3 = 6 Problem 3 :
x ⇾ -1 x + 1 x 2 + 2x + 1 Solution :
Given , x ⇾ -1 x + 1 x 2 + 2x + 1 = x ⇾ -1 x + 1 ( x + 1 ) 2 = x ⇾ -1 1 x + 1 = 1 - 1 + 1 = 1 0 Does not exist.
Problem 4 :
x ⇾ 3 x 2 - 3x x 2 - 4x + 3 Solution :
Given , x ⇾ 3 x 2 - 3x x 2 - 4x + 3 = x ⇾ 3 x ( x - 3 ) ( x - 1 ) ( x - 3 ) = x ⇾ 3 x x - 1 = 3 3 - 1 = 3 2 Problem 5 :
x ⇾ 7 x - 7 x 2 - 49 Solution :
Given , x ⇾ 7 x - 7 x 2 - 49 = x ⇾ 7 x - 7 x 2 - 7 2 = x ⇾ 7 x - 7 ( x + 7 ) ( x - 7 ) = x ⇾ 7 1 x + 7 = 1 7 + 7 = 1 14 Problem 6 :
x ⇾ 2 x + 2 x 3 + 8 Solution :
Given , x ⇾ 2 x + 2 x 3 + 8 = x ⇾ 2 x + 2 x 3 + 2 3 = 2 + 2 2 3 + 8 = 2 + 2 8 + 8 = 4 16 = 1 4 Problem 7 :
x ⇾ 2 2x + 4 6 + 3x Solution :
Given , x ⇾ 2 2x + 4 6 + 3x = x ⇾ 2 2 ( x + 2 ) 3 ( 2 + x ) = x ⇾ 2 = 2 3 Problem 8 :
x ⇾ 0 2 x - 1 2 2x - 1 Solution :
Given , x ⇾ 0 2 x - 1 2 2x - 1 = x ⇾ 0 2 x - 1 2 x 2 - 1 = x ⇾ 0 2 x - 1 2 x 2 - 1 2 = x ⇾ 0 2 x - 1 2 x - 1 2 x + 1 = x ⇾ 0 1 2 x + 1 = 1 2 0 + 1 = 1 1 + 1 = 1 2 Problem 9 :
x ⇾ 2 3x 2 - 4x - 4 2x 2 - 8 Solution :
Given , x ⇾ 2 3x 2 - 4x - 4 2x 2 - 8 = x ⇾ 2 ( x - 2 ) ( 3x + 2 ) 2 x 2 - 4 = x ⇾ 2 ( x - 2 ) ( 3x + 2 ) 2 x 2 - 2 2 = x ⇾ 2 ( x - 2 ) ( 3x + 2 ) 2 x + 2 ( x - 2 ) = x ⇾ 2 3x + 2 2 ( x + 2 ) = 3 ( 2 ) + 2 2 ( 2 + 2 ) = 8 8 = 1 Problem 10 :
x ⇾ 2 x 2 + x - 6 x - 2 Solution :
Given , x ⇾ 2 x 2 + x - 6 x - 2 = x ⇾ 2 ( x + 3 ) ( x - 2 ) x - 2 = x ⇾ 2 x + 3 = 2 + 3 = 5 Problem 11 :
x ⇾ 8 x 3 - 64x x - 8 Solution :
Given , x ⇾ 8 x 3 - 64x x - 8 = x ⇾ 8 x x 2 - 64 x - 8 = x ⇾ 8 x x 2 - 8 2 x - 8 = x ⇾ 8 x ( x + 8 ) ( x - 8 ) x - 8 = x ⇾ 8 x ( x + 8 ) = 8 ( 8 + 8 ) = 8 ( 16 ) = 128 Problem 12 :
x ⇾ 3 3x - 9 12 - 4x Solution :
Given , x ⇾ 3 3x - 9 12 - 4x = x ⇾ 3 3 ( x - 3 ) 4 ( 3 - x ) = x ⇾ 3 - 3 ( 3 - x ) 4 ( 3 - x ) = x ⇾ 3 - 3 4 = - 3 4 Problem 13 :
x ⇾ 3 x 3 - 27 x - 3 Solution :
Given , x ⇾ 3 x 3 - 27 x - 3 = x ⇾ 3 x 3 - 3 3 x - 3 = x ⇾ 3 ( x - 3 ) x 2 + 3x + 9 x - 3 = x ⇾ 3 x 2 + 3x + 9 = 3 2 + 3 ( 3 ) + 9 = 9 + 9 + 9 = 27
