Đáp án:
$\lim _{x\to \infty }\left(\dfrac{\sqrt{x^2-3x}}{\:x+2}\right)$
$=\lim _{x\to \infty \:}\left(\dfrac{\dfrac{\sqrt{x^2-3x}}{x}}{1+\dfrac{2}{x}}\right)$
$=\dfrac{\lim _{x\to \infty \:}\left(\dfrac{\sqrt{x^2-3x}}{x}\right)}{\lim _{x\to \infty \:}\left(1+\dfrac{2}{x}\right)}$
$=1$
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$\lim _{x\to -\infty }\left(\dfrac{\sqrt{x^2-3x}}{\:x+2}\right)$
$=\lim _{x\to \:-\infty \:}\left(\dfrac{\dfrac{\sqrt{x^2-3x}}{x}}{1+d\frac{2}{x}}\right)$
$=\dfrac{\lim _{x\to \:-\infty \:}\left(\dfrac{\sqrt{x^2-3x}}{x}\right)}{\lim _{x\to \:-\infty \:}\left(1+\dfrac{2}{x}\right)}$
$=-1$
