Question

Limite no infinito


$\lim_{n \to \infty} \frac{ \sqrt{4x^-25} }{2x-5}$

Answer 1

$\lim_{x \to \infty} \frac{ \sqrt{4x^-25} }{2x-5}$


$\lim_{x \to \infty} \frac{ \sqrt{ \frac{4x^2}{x^2} - \frac{25}{x^2} } }{ \frac{2x}{x}- \frac{5}{x} }$


$\lim_{x \to \infty} \frac{ \sqrt{ \frac{4\not x^2}{\not x^2} - \frac{\not 25}{\not x^2} } \to 0 }{ \frac{2\not x}{\not x}- \frac{\not5}{\not x} \to 0 }$


$\lim_{x \to \infty} \frac{ \sqrt{4} }{2}$


$\lim_{x \to \infty} \frac{2}{2} = 1$



$\boxed {\lim_{x \to \infty} \frac{ \sqrt{4x^-25} }{2x-5} = 1}$