Question

Bantu jawab dong, pleaseee ... bantu yg bisaa

Answer 1

1)
lim(x->2) { 4 - x² } (2 + √(2x)} / { 2- √(2x)}{2 +√(2x)}
lim(x->2) { (2-x)(2+x)(2+√(2x)} / (4 - 2x)
lim (x->2) { (2-x)(2+x)(2+√(2x)}/ {2(2-x)}
lim(x->2)  { (2+x)(2+√(2x)} /(2)
x = 2 --> { (2+2)(2+√4)} / 2 = (4)(4)/2 = 4

2) bagikan dengan masing masing dengan x⁷
limit =- 5/0 = ~

3) limit (x->0)  tan 5x/3x = 5/3

4) lim (x->0) sin (x-1) / (x² -1) =
lim (x->0)  sin (x-1)./ (x+1)/(x-1) = 1/(x+1)
x = 0 --> limit = 1/(0+1)= 1