Question

o Valor de lim 3x+2............................................................................................................

Answer 1

Olá

Alternativa correta, letra e) 0

$\displaystyle \lim_{x \to -\infty}~ \frac{3x+2}{x^2-5x+6} \\\\\\\text{Poe o termo 'x' com maior indice (grau) em evidencia}\\\\\\ \lim_{x \to -\infty}~ \frac{x(3+ \frac{2}{x} )}{x^2(1- \frac{5x}{x^2} + \frac{6}{x} )} \\\\\\\text{Simplifica}\\\\\\ \lim_{x \to -\infty}~ \frac{\diagup\!\!\!\!x(3+ \frac{2}{x} )}{x^{\diagup\!\!\!\!2}(1- \frac{5\diagup\!\!\!\!\!x}{x^{\diagup\!\!\!\!\!2}} + \frac{6}{x} )}$

$\displaystyle \lim_{x \to -\infty}~ \frac{3+ \frac{2}{x} }{x(1- \frac{5}{x} + \frac{6}{x} )} \\\\\\\text{Propriedade de limite quando tende ao }\infty\\\\\\ \lim_{x \to \pm\infty} ~ \frac{k}{x} ~=~ \frac{k}{\pm\infty} ~=~0~~~~ ~~ ~~~~ ~k\in R\\\\\\\text{Aplicando essa prorpriedade}\\\\\\\lim_{x \to -\infty}~ \frac{3+ \diagup\!\!\!\!\!\frac{2}{x}^~0 }{x(1-{( \diagup\!\!\!\!\frac{5}{x})^0} + (\diagup\!\!\!\!\frac{6}{x})^0 )}$

$\displaystyle \lim_{x \to -\infty}~ \frac{3}{x\cdot 1} ~=~ \frac{3}{-\infty \cdot 1} ~=~ \frac{3}{-\infty} ~=~\boxed{0}$