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Jawab2. substitusi x= 3
limit = √3²-4 = √5
3. substitusi x= 2
( - 4(2)² +6(2) - 8 ) /(3 (2)² +7(2) - 3) =
= (- 16 + 12 -8) / (12 +14 - 3)
= (-12)/(23)
4. lim(x->4) {(√x - 2) /(√x + 2)(√x - 2) }
limit(x -> 4) (1/ √x + 2)
x = 4 -->limit = 1/ √4 + 2 = 1/ (2 + 2) = 1/4
5. limit (x -> 7) {(2 - √(x-3))(2 + √(x -3)) } / (x-7)(x+7)
limit (x -> 7) ( 4 - (x - 3)) / (x - 7)(x + 7)
limit(x -> 7 ( 7 - x) / (x-7)(x +7)
limit(x->7) - (x - 7) / (x -7)(x+7) = 1/(x+7)
x= 7 --> limit = 1/(7+7) = 1/14