Question

resolva:

lim x -> infinito

raiz de 2x^2 + 1 / 3 x - 5

Answer 1

Olá


$\displaystyle \mathsf{ \lim_{x \to \infty} ~ \frac{ \sqrt{2x^2+1} }{3x-5} }\\\\\\\text{Limites envolvendo o infinito, e bem simples, basta colocar o 'x'}\\\text{com maior indice em evidencia}\\\\\\ \mathsf{ \lim_{x \to \infty}~ \frac{ \sqrt{x^2(2+ \frac{1}{x^2} )} }{x(3- \frac{5}{x} )} }\\\\\\\text{O x}^2~\text{Sai da raiz com o modulo}\\\\\\\mathsf{ \lim_{x \to \infty} ~ \frac{ |x|\sqrt{(2+ \frac{1}{x^2} )} }{x(3- \frac{5}{x} )} }\\\\\text{Como se trata de uma raiz quadrada, o 'x' sai da raiz positivo}$

$\displaystyle \mathsf{ \lim_{x \to \infty} ~ \frac{ x\cdot \sqrt{(2+ \frac{1}{x^2} )} }{x(3- \frac{5}{x} )} }\\\\\\\text{Simplifica}\\\\\\\mathsf{ \lim_{x \to \infty} ~ \frac{ \diagup\!\!\!\!x\cdot \sqrt{(2+ \frac{1}{x^2} )} }{\diagup\!\!\!\!x(3- \frac{5}{x} )} }\\\\\\ \mathsf{ \lim_{x \to \infty} ~ \frac{ \sqrt{2+ \frac{1}{x^2} } }{3- \frac{5}{x} } }\\\\\\\text{Pelas propriedade de limites, sabemos que}\\\\\\\mathsf{ \lim_{x \to \pm\infty} ~ \frac{k}{x}~=~ \frac{k}{\pm\infty}~=~0~~~~~ ~~~ ~k\in\Re }$

$\displaystyle \text{Resolvendo o limite}\\\\\\\mathsf{ \lim_{x \to \infty} ~ \frac{ \sqrt{2+ \frac{1}{x^2} } }{3- \frac{5}{x} } ~=~ \frac{ \sqrt{2+ \frac{1}{\infty} } }{3- \frac{5}{\infty} } ~=~ \frac{ \sqrt{2+0} }{3-0}~=~ \boxed{\frac{ \sqrt{2} }{3} }}$