Question

Calcule o limite, se existir: \lim _{x\to \infty} \frac{\sqrt[]{x^2 1}}{x 1}

Answer 1

$\displaystyle L=\lim\limits _{x\to \infty} \frac{\sqrt{x^2+1}}{x+1}\\ \\ L=\lim\limits _{x\to \infty} \frac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}\\ \\ L=\dfrac{\sqrt{1+0}}{1+0}\\ \\ \\ \boxed{\boxed{L=1}}$