Question

1. lim┬(x→0)⁡〖〖(√(x^2+ 4)-2)/x〗^ 〗
2. lim┬(x→10)⁡〖〖(√(x-1)-3)/(x-10)〗^ 〗
3.〖lim⁡〗┬(x→5)⁡〖〖(25-x^2)/(5-√5x)〗^ 〗
4. lim┬(x→2)⁡〖〖(4-√8x)/(√(8-2x)-2)〗^ 〗

Tolong bantuin ya... pake cara

Answer 1

1.
= lim x->0 √(x²+4) - 2 / x . √(x²+4)+2 / √(x²+4)+2
= lim x->0  x²+4-4 / x(√(x²+4)+2)
= lim x->0  x(x) / x((√x²+4)+2)
x nya dicoret
= lim x->0 x / (√x²+4)+2
= 0 / √4 + 2
= 0 / 2+2
= 0 / 4
= 0
2.
= lim x->10  √(x-1) - 3 / x-10 . √(x-1) + 3 / √(x-1) + 3
= lim x->10  x-1-9 / x-10(√(x-1)+3)
= lim x->10  x-10 / x-10 (√(x-1) + 3)
coret x-10
= lim x->10  1 / √(x-1) + 3
= 1 / √(10-1) + 3
= 1 / √9 + 3
= 1/ 3+3
= 1/6
3.
= lim x->5  25-x² / 5-√(5x) . 5+√(5x) / 5+√(5x)
= lim x->5  (25-x²)(5+√(5x) / 25-5x
= lim x->5  25(-x²)(5+√(5x)) / 25(-5x)
coret 25
= lim x->5  -x²(5+√(5x)) / -5x
= -(5²) (5+√(5(5)) / -5(5)
= 25 (5+√25) / -25
= 25(5+5) / -25
= 25(10) / -25
= 250/-25
= -10
4.
= lim x->2  4-8x / √(8-2x) - 2 . √(8-2x) + 2 / √(8-2x) + 2
= lim x->2  (4-8x)(√(8-2x)+2) / 8-2x-4
= lim x->2  (4-8x)(√(8-2x)+2) / 4-2x
= lim x->2  4(1-2x)(√(8-2x)+2) / 4(-2x)
coret 4
= lim x->2  (1-2x)(√(8-2x)+2) / (-2x)
= (1-2(2))(√(8-2(2))+2) / (-2(2))
= (1-4)(√(8-4)+2) / -4
= -3(√4 + 2) / -4
= -3(2+2) / -4
= -3(4) / -4
= 3

Maaf, kalau ada yang salah.