Answer:
1/3
Step-by-step explanation:
[tex][x^2 - ln(2/x)]/[3x^2 + 2x]\\[/tex]
∞/∞
[tex]\frac{d}{dx}[x^2-ln(2/x)] =2x+\frac{1}{x}\\ \frac{d}{dx}[3x^2+2x]=6x+2\\\frac{2x+\frac{1}{x}}{6x+2}[/tex]
Also ∞/∞
[tex]\frac{d}{dx}[2x+\frac{1}{x}]=2-1/x^2\\\frac{d}{dx}[6x+2]=6\\\\[/tex]
[tex]\lim_{x \to \infty} \frac{2-\frac{1}{x^2} }{6}\\=\frac{2}{6} \\=\frac{1}{3}[/tex]
