Mohoh Bantuannya #NoSPam

27) Lim (√(5x + 4) - √(3x + 9)) / 4x
= Lim [(√(5x + 4))/x - (√(3x + 9))/x] / (4x)/x
= Lim [√(5x/x^2 + 4/x^2) - √(3x/x^2 + 9/x^2)] / 4
= Lim [√(5/x + 4/x^2) - √(3/x + 9/x^2)] / 4
= [√(0 + 0) - √(0 + 0)] / 4
= 0/4
= 0

28) Lim [√(x + 2) - √(x - 1)]
= Lim [√(x + 2) - √(x - 1)] . [√(x + 2) + √(x - 1)]/[√(x + 2) + √(x - 1)]
= Lim [(x + 2) - (x - 1)] / [√(x + 2) + √(x - 1)]
= Lim 3 / [√(x + 2) + √(x - 1)] . (1/√x)/(1/√x)
= Lim (3/√x) / [√(x/x + 2/x) + √(x/x - 1/x)]
= 0/(√(1 + 0) + √(1 - 0)]
= 0/2
= 0

Bab Limit tak hingga
Matematika SMA Kelas XI

27]
Lim (√(5x + 4) - √(3x + 9)) / 4x
x→~

= Lim (√((5x + 4)/x²) - √((3x + 9)/x²))/(4x / x)
...x→~

= Lim (√(5/x + 4/x²) - √(3/x + 9/x²))/4
...x→~

※※※ a/~ = 0

(√0 - √0)/4
= 0/4
= 0

28]
Lim (√(x + 2) - √(x - 1))
x→~

= Lim (√(x/x + 2/x) - √(x/x - 1/x)
...x→~

= √1 - √1
= 0