Question

lim(x tinde la infinit) (4x-ln(x^2+1)) ?? Heeelp

Answer 1

$l =\lim\limits_{x\to \infty} \left[4x-\ln(x^2+1)\right] =\lim\limits_{x\to \infty} \left[4x-\ln(x^2)\right] =$

$=\lim\limits_{x\to \infty} (4x-2\ln x)\\\\\ln x = t \Rightarrow x = e^t$

$x \to \infty \Rightarrow t\to \infty\\ \\ l =\lim\limits_{t\to \infty}(4e^t - 2t) =\infty$