Question

calcular estes limites:
$\lim_{x \to \infty} ln \frac{x}{x+1}$
$\lim_{x \to \infty} [ln(2x+1)-ln(x+3)]$

Answer 1

Primeira:

$\lim_{x\to\infty}\ln\frac{x}{x+1}=\\\\\ln[\lim_{x\to\infty}\frac{x}{x(1+\frac{1}{x})}]=\\\\\ln[\frac{1}{1+0}]=\\\\\ln1=\\\\\boxed{0}$

Segunda:

$\lim_{x\to\infty}\ln\frac{2x+1}{x+3}=\\\\\ln[\lim_{x\to\infty}\frac{x(2+\frac{1}{x})}{x(1+\frac{3}{x})}]=\\\\\ln[\lim_{x\to\infty}\frac{2+\frac{1}{x}}{1+\frac{3}{x}}]=\\\\\ln(\frac{2+0}{1+0})=\\\\\boxed{\ln2}$