Matematika Sekolah Menengah Atas tolong jawabannya ya tepat dongg​

tolong jawabannya ya tepat dongg​

[tex]\lim_{x \to \infty} \frac{\sqrt{12x^2-5x} +3x}{\sqrt{8x^2+4} }[/tex]

[tex]= \lim_{x \to \infty} \frac{\sqrt{12x^2-5x} +\sqrt{(3x)^2} }{\sqrt{8x^2+4} }[/tex]

[tex]= \lim_{x \to \infty} \frac{\sqrt{12x^2-5x} +\sqrt{9x^2} }{\sqrt{8x^2+4} }[/tex]

[tex]= \lim_{x \to \infty} \frac{\sqrt{\frac{12x^2}{x^2} -\frac{5x}{x^2} } +\sqrt{\frac{9x^2}{x^2} } }{\sqrt{\frac{8x^2}{x^2}+\frac{4}{x^2} } }[/tex]

[tex]= \lim_{x \to \infty} \frac{\sqrt{12-\frac{5}{x} } +\sqrt{9} }{\sqrt{8+\frac{4}{x^2} } }[/tex]

[tex]= \frac{\sqrt{12-\frac{5}{\infty} } +\sqrt{9} }{\sqrt{8+\frac{4}{\infty^2} } }[/tex]

[tex]= \frac{\sqrt{12-0 } +\sqrt{9} }{\sqrt{8+0 } }[/tex]

[tex]= \frac{\sqrt{12} +\sqrt{9} }{\sqrt{8} }[/tex]

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

Jawaban : A

[answer.2.content]