Tìm giới hạn 1. lim [(-2x^2+x-1) / (3+x)] x–>-vô cực 2. lim [(3x^2+x+5) / (1+2x)] x–>-vô cực

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1. 1.

   $\lim\limits_{x\to -\infty}\dfrac{-2x^2+x-1}{x+3}$

   $=\lim\limits_{x\to -\infty}x.\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{1+\dfrac{3}{x}}$

   $=+\infty$

   2.

   $\lim\limits_{x\to -\infty}\dfrac{3x^2+x+5}{2x+1}$

   $=\lim\limits_{x\to -\infty}x.\dfrac{3+\dfrac{1}{x}+\dfrac{5}{x^2}}{2+\dfrac{1}{x}}$

   $=-\infty$

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2. $1)$ $\lim\limits_{x\to -∞} \dfrac{-2x^2+x-1}{3+x}$

   $=\lim\limits_{x\to -∞} \dfrac{x^2 (-2+\dfrac{1}{x}+\dfrac{1}{x^2})}{x(\dfrac{3}{x}+1)}$

   $=\lim\limits_{x\to -∞} x .\dfrac{-2+\dfrac{1}{x}+\dfrac{1}{x^2}}{\dfrac{3}{x}+1}=+∞$

   (vì $\lim\limits_{x\to -∞} x=-∞$

   và $\lim\limits_{x\to -∞} \dfrac{-2+\dfrac{1}{x}+\dfrac{1}{x^2}}{\dfrac{3}{x}+1}=-2<0$)

   $\\$

   $2)$ $\lim\limits_{x\to -∞} \dfrac{3x^2+x+5}{1+2x}$

   $=\lim\limits_{x\to -∞} \dfrac{x^2 (3+\dfrac{1}{x}+\dfrac{5}{x^2})}{x(\dfrac{1}{x}+2)}$

   $=\lim\limits_{x\to -∞} x .\dfrac{3+\dfrac{1}{x}+\dfrac{5}{x^2}}{\dfrac{1}{x}+2}=-∞$

   (Vì $\lim\limits_{x\to -∞} x=-∞$

   và $\lim\limits_{x\to -∞} \dfrac{3+\dfrac{1}{x}+\dfrac{5}{x^2}}{\dfrac{1}{x}+2}=\dfrac{3}{2}>0$)

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