Tính lim( x–>âm vô cùng ) (căn x^2 +3x-2 )+x +4

Ta có

$\underset{x \to -\infty}{\lim} (\sqrt{x^2 + 3x – 2} + x + 4) = \underset{x \to -\infty}{\lim} \dfrac{x^2 + 3x – 2 – (x^2 + 8x + 16)}{\sqrt{x^2 + 3x – 2} – x – 4}$

$= \underset{x \to -\infty}{\lim} \dfrac{-5x -18}{\sqrt{x^2 + 3x – 2} – x – 4}$

$= \underset{x \to -\infty}{\lim} \dfrac{5 + \frac{18}{x}}{\sqrt{x^2 + \frac{3}{x} – \frac{2}{x^2}} + 1 + \frac{4}{x}}$

$= \dfrac{5}{1 + 1} = \dfrac{5}{2}$

Vậy

$\underset{x \to -\infty}{\lim} (\sqrt{x^2 + 3x – 2} + x + 4) = \dfrac{5}{2}$.