mohon dibantu yah..... please soalnya besok dikumpulin

jawab


2. 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