Question

valor do lim x ----> +∞ ( 5x^3-6x+1/6x^5+x+3) está expresso na alternativa:

a) -∞

b) 1

c) -1

d) 0

e) +∞

Answer 1

Olá

Alternativa correta, letra D)

$\displaystyle \lim_{x \to \infty} ~ \frac{5x^3-6x+1}{6x^5+x+3}$

Põe o termo X com maior grau em evidência

$\displaystyle \lim_{x \to \infty} ~ \frac{x^3(5- \frac{6x}{x^3} + \frac{1}{x^3} )}{x^5(6+ \frac{x}{x^5} + \frac{3}{x^5}) } \\ \\ \\ \text{Simplifica} \\ \\ \\ \lim_{x \to \infty} ~ \frac{x^{\diagup\!\!\!\!3}(5- \frac{6\diagup\!\!\!\!x}{x^{\diagup\!\!\!\!3}} + \frac{1}{x^3} )}{x^{\diagup\!\!\!\!5}(6+ \frac{\diagup\!\!\!\!x}{x^{\diagup\!\!\!\!5}} + \frac{3}{x^5}) } \\ \\ \\ \lim_{x \to \infty} ~ \frac{(5- \frac{6}{x^2} + \frac{1}{x^3} )}{x^2(6+ \frac{1}{x^4} + \frac{3}{x^5}) }$

Pelas propriedades de limites, sabemos que:

$\displaystyle \frac{k}{\pm \infty} ~=~0~~~~~~\text{com k}\in R$

$\displaystyle \lim_{x \to \infty} ~ \frac{(5- \frac{6}{x^2} + \frac{1}{x^3} )}{x^2(6+ \frac{1}{x^4} + \frac{3}{x^5}) }=\frac{(5- \frac{6}{\infty^2} + \frac{1}{\infty^3} )}{\infty^2(6+ \frac{1}{\infty^4} + \frac{3}{\infty^5}) }= \frac{5-0+0}{\infty(6+0+0)} \\ \\ \\ = \frac{5}{\infty\cdot6} = \frac{5}{\infty} =\boxed{0}$