lim┬(x→0)⁡〖3x/((√(9+x))-(√(9-x)))

jawabannya 9. cara ada di gambar berikut

[tex] \lim_{x \to 0} \frac{3x}{ \sqrt{9+x}\,\,-\sqrt{9-x}} [/tex]

[tex]= \lim_{x \to 0} \frac{3x}{ \sqrt{9+x}\,\,-\sqrt{9-x}} \times \frac{\sqrt{9+x}\,\,+\sqrt{9-x}}{\sqrt{9+x}\,\,+\sqrt{9-x}} [/tex]

[tex]= \lim_{x \to 0} \frac{(3x)(\sqrt{9+x}\,\,+\sqrt{9-x})}{(9+x)\,\,-(9-x)}[/tex]

[tex]= \lim_{x \to 0} \frac{(3x)(\sqrt{9+x}\,\,+\sqrt{9-x})}{9+x-9+x}[/tex]

[tex]= \lim_{x \to 0} \frac{(3x)(\sqrt{9+x}\,\,+\sqrt{9-x})}{2x}[/tex]

[tex]= \lim_{x \to 0} \frac{3(\sqrt{9+x}\,\,+\sqrt{9-x})}{2}[/tex]

[tex]= \frac{3(\sqrt{9+0}\,\,+\sqrt{9-0})}{2}[/tex]

[tex]= \frac{3(\sqrt{9}\,\,+\sqrt{9})}{2}[/tex]

[tex]= \frac{3(3+3)}{2}[/tex]

[tex]= \frac{3(6)}{2}[/tex]

[tex]= \frac{18}{2}[/tex]

[tex]= 9[/tex]