tolong di jawab ya kakak

Lim (3x) / (√(9 + x) - √(9 - x)) . (√(9 + x) + √(9 - x))/(√(9 + x) + √(9 - x))
x=>0

Lim 3x (√(9 + x) + √(9 - x)) / ((9 + x) - (9 - x))
x=>0

Lim 3x (√(9 + x) + √(9 - x)) / (2x)
x=>0

Lim 3/2 (√(9 + x) + √(9 - x))
x=>0

= 3/2 (√(9 + 0) + √(9 - 0)

= 3/2 (3 + 3)

= 9

[tex]\displaystyle \lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}\cdot\frac{\sqrt{9+x}+\sqrt{9-x}}{\sqrt{9+x}+\sqrt{9-x}}\\\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\lim_{x\to0}\frac{3x(\sqrt{9+x}+\sqrt{9-x})}{9+x-9+x}\\\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\lim_{x\to0}\frac{3x(\sqrt{9+x}+\sqrt{9-x})}{2x}\\\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\lim_{x\to0}\frac{3(\sqrt{9+x}+\sqrt{9-x})}{2}[/tex]
[tex]\displaystyle \lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\frac{3(\sqrt{9+0}+\sqrt{9-0})}{2}\\\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=\frac{3(3+3)}{2}\\\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=3(3)\\\boxed{\boxed{\lim_{x\to0}\frac{3x}{\sqrt{9+x}-\sqrt{9-x}}=9}}[/tex]