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Do $x \to+\infty$ nên ta có $1-2x <0$ suy ra
$L=\mathop {\lim }\limits_{x \to+\infty}\left(1-2x\right)\sqrt{\dfrac{3x-11}{x^3+1}}=-\mathop {\lim }\limits_{x \to+\infty}\sqrt{\dfrac{(3x-11)\left(1-2x\right)^2}{x^3+1}}$
$L=-\mathop {\lim }\limits_{x \to+\infty}\sqrt{\dfrac{(3-\dfrac{11}{x})\left(\dfrac{1}{x}-2\right)^2}{1+\dfrac{1}{x^3}}}=-\sqrt{\dfrac{3.4}{1}}=-3$
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