Tính lim( x–>âm vô cùng ) (căn x^2 +3x-2 )+x +4 25/11/2021 Bởi Ximena 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 $\underset{x \to -\infty}{\lim} (\sqrt{x^2 + 3x – 2} + x + 4) = \dfrac{5}{2}$. Bình luận
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
$\underset{x \to -\infty}{\lim} (\sqrt{x^2 + 3x – 2} + x + 4) = \dfrac{5}{2}$.
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