lim x➡3 x-√2x+3/x^2-9 a.0 b.1/9 c.1/6 d.8 e.9

= (x^2 - (2x+3)) / (x^2 -9)(x+akar(2x+3))
= (x^2 - 2x - 3)/(x^2 - 9)(x+akar(2x+3))
= (x-3)(x+1)/(x-3)(x+3)(x+akar(2x+3))
= (x+1)/(x+3)(x+akar(2x+3))
= (3+1)/(3+3)(3+akar(2×3 + 3)
= 4/6(3+akar9)
= 2/3(3+3)
= 2/3×6
= 2/18
= 1/9


smg membantu