Tìm các giới hạn a, lim [(3x^2 – 2x – 1) / (x^3 – 1)] x–>1 b, lim [{căn(1+2x)} – {căn3 (1+3x)}] / [x^2] x–>0 c, lim (x+3) / (x-3) x–>3-

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1. a,

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

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

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

   $=\dfrac{3.1+1}{1^2+1+1}=\dfrac{4}{3}$

   b,

   $\lim\limits_{x\to 0}\dfrac{\sqrt{1+2x}-(x+1)+(x+1)-\sqrt[3]{1+3x}}{x^2}$

   $=\lim\limits_{x\to 0}\dfrac{ \dfrac{1+2x-x^2-2x-1}{ \sqrt{1+2x}+x+1}+\dfrac{x^3+3x^2+3x+1-1-3x}{(x+1)^2+(x+1)\sqrt[3]{1+3x}+\sqrt[3]{1+3x}^2} }{x^2}$

   $=\lim\limits_{x\to 0}\dfrac{-1}{\sqrt{1+2x}+x+1}+\dfrac{x+3}{(x+1)^2+(x+1)\sqrt[3]{1+3x}+\sqrt[3]{1+3x}^2}$

   $=\dfrac{-1}{1+1}+\dfrac{0+3}{1+1.1+1}$

   $=\dfrac{1}{2}$

   c,

   $\lim\limits_{x\to 3^-}(x+3)=3+3=6>0$

   $\lim\limits_{x\to 3^-}(x-3)=0$

   $\Rightarrow\lim\limits_{x\to 3^-}\dfrac{x+3}{x-3}=-\infty$

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