Difference of two infinities

1.- $\lim_{x \to -\infty} \left(\sqrt{x^2 + 4x + 1} - \ln (-x)\right) =$ (make the substitution u = - x), $= \lim_{u \to \infty} \left(\sqrt{u^2 - 4u + 1} - \ln (u)\right) = \lim_{u \to \infty} \frac{(\sqrt{u^2 - 4u + 1} - \ln (u))(\sqrt{u^2 - 4u + 1} + \ln (u))}{\sqrt{u^2 - 4u + 1} + \ln (u)} =$ $\lim_{u \to \infty} \frac{u^2 - 4u + 1 - \ln^2 (u)}{\sqrt{u^2 - 4u + 1} + \ln (u)} =$ (Divide numerator and denominator by u) $= \lim_{u \to \infty} \frac{u - 4 + \frac{1}{u} - \frac{\ln^2 (u)}{u}}{\sqrt{1 - \frac{4}{u} + \frac{1}{u^2}} + \frac{\ln (u)}{u}} = + \infty$ due to $\displaystyle \lim_{u \to \infty} \frac{\ln^2 (u)}{u} = \lim_{u \to \infty} \frac{\ln (u)}{u} = 0$

2.- $\lim_{x \to -\infty} \left(\sqrt{x^2 + 4x + 1} - \ln (-x)\right) = \lim_{x \to \infty} \left(\sqrt{x^2 - 4x + 1} - \ln (x)\right) = (*)$ (It can be proved that $\quad \lim_{x \to \infty} (\sqrt{x^2 - 4x +1} - (x - 2)) = 0 \Rightarrow$ ) $(*) = \lim_{x \to -\infty} \left(\sqrt{x^2 - 4x + 1} - (x - 2) + (x - 2) - \ln (x)\right) =$ $= \lim_{x \to \infty} (\sqrt{x^2 - 4x +1} - (x - 2)) + \lim_{x \to \infty} ((x - 2) - \ln (x)) = +\infty$ due to the first limit in the sum above is equal to 0 and the second limit above is equal to $\infty$ .