tìm giới hạn lim (x->1) $\frac{\sqrt[]{x+3}-\sqrt[]{3x+1}}{x-1}$

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1. Đáp án:

   $\begin{array}{l}
   \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3}  – \sqrt {3x + 1} }}{{x – 1}}\\
    = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {x + 3}  – \sqrt {3x + 1} } \right)\left( {\sqrt {x + 3}  + \sqrt {3x + 1} } \right)}}{{\left( {x – 1} \right).\left( {\sqrt {x + 3}  + \sqrt {3x + 1} } \right)}}\\
    = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3 – \left( {3x + 1} \right)}}{{\left( {x – 1} \right)\left( {\sqrt {x + 3}  + \sqrt {3x + 1} } \right)}}\\
    = \mathop {\lim }\limits_{x \to 1} \dfrac{{2 – 2x}}{{\left( {x – 1} \right)\left( {\sqrt {x + 3}  + \sqrt {3x + 1} } \right)}}\\
    = \mathop {\lim }\limits_{x \to 1} \dfrac{{ – 1}}{{\left( {\sqrt {x + 3}  + \sqrt {3x + 1} } \right)}}\\
    = \dfrac{{ – 1}}{{\sqrt {1 + 3}  + \sqrt {3 + 1} }}\\
    =  – \dfrac{1}{4}
   \end{array}$

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2. $\lim\limits_{x\to 1}\dfrac{\sqrt{x+3}-\sqrt{3x+1}}{x-1}$

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

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

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

   $=\dfrac{-2}{2+2}$

   $=\dfrac{-1}{2}$

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