Question

lim ln (9x) - ln (3x) com x--> oo

pf alguem que saiba resolver estes limites de ln? (os mais dificeis)

Answer 1

a,b,c é só observar o grafico da funçao log


d)

$\lim_{x \to \infty} \; ln( \frac{x}{x+1} )\\\\\text{coloca x em evidencia no denominador}\\\\ \lim_{x \to \infty} \; ln( \frac{x}{x(1+ \frac{1}{x}) } )\\\\ = \lim_{x \to \infty} \; ln( \frac{1}{1+ \frac{1}{x} } ) = ln( \frac{1}{1+ \frac{1}{\infty} }) = ln( \frac{1}{1+0} ) = ln(1)=0$


e)
$\lim_{x \to \infty} ln(2x+1)-ln(x+3)\\\\\text{aplica a propriedade de log}\;\; \boxed{\boxed{ln(a)-ln(b)= \frac{ln(a)}{ln(b)} }}\\\\ \lim_{x \to \infty} ln(\frac{2x+1}{x+3}) \\\\\text{coloca x em evidencia (divide o numerador e o denominador por x)}\\\\ \lim_{x \to \infty} ln( \frac{x(2+ \frac{1}{x} )}{x(1+ \frac{3}{x}) } ) = \lim_{x \to \infty} ln( \frac{2+ \frac{1}{x} }{1+ \frac{3}{x} } )= ln( \frac{2+0}{1+0} ) = ln(2)$

f) mesma coisa , lembrando que a²-b² = (a-b)(a+b)

g) 
$\lim_{x \to \infty} xln(2)- ln(3^x+1) \\\\ \text{aplicando a propriedade } \;\; a*ln(b)= ln(b^a)\\\\\\ \lim_{x \to \infty} ln(2^x)- ln(3^x+1) \\\\ = \lim_{x \to \infty} ln( \frac{2^x}{3^x+1}) \\\\ = \lim_{x \to \infty} ln( \frac{2^x}{3^x(1+ \frac{1}{3^x}) }) \\\\ = \lim_{x \to \infty} ln\left ( (\frac{2}{3})^x* \frac{1}{(1+ \frac{1}{3^x}) }\right ) \\\\ = ln \left ( \lim_{x \to \infty}( \frac{2}{3} )^x * \lim_{x \to \infty} \frac{1}{1+ \frac{1}{3^x} } \right)$


$\text{resolvendo}\\\\ \lim_{x \to \infty}( \frac{2}{3} )^x = 0 \; porque \; 3\ \textgreater \ 2 \\\\ \lim_{x \to \infty} \frac{1}{1+ \frac{1}{3^x} } \right) = \frac{1}{1+ \frac{1}{\infty} } = \frac{1}{1+ 0 } = 1\\\\\text{temos}\\\\ ln \left ( \lim_{x \to \infty}( \frac{2}{3} )^x * \lim_{x \to \infty} \frac{1}{1+ \frac{1}{3^x} } \right) = ln(0*1) = ln(0)\approx -\infty$