Question

Use calculus to evaluate (lhopitals rule)
lim x>infinity x^2 - ln(2/x)/3x^2 + 2x

Answer 1

Answer:

1/3

Step-by-step explanation:

$[x^2 - ln(2/x)]/[3x^2 + 2x]$

∞/∞

$\frac{d}{dx}[x^2-ln(2/x)] =2x+\frac{1}{x}\\ \frac{d}{dx}[3x^2+2x]=6x+2\\\frac{2x+\frac{1}{x}}{6x+2}$

Also ∞/∞

$\frac{d}{dx}[2x+\frac{1}{x}]=2-1/x^2\\\frac{d}{dx}[6x+2]=6$

$\lim_{x \to \infty} \frac{2-\frac{1}{x^2} }{6}\\=\frac{2}{6} \\=\frac{1}{3}$