lim (x->0) $\frac{\sqrt[5]{5x+1}-1}{x}$ lim (x->8) $\frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2}$

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1. Đáp án:

   $\begin{array}{l}
   \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[5]{{5x + 1}} – 1}}{x}\\
    = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {5x + 1 – 1} \right)}}{{x.\left( {{{\left( {\sqrt[5]{{5x + 1}}} \right)}^4} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^3} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^2} + \sqrt[5]{{5x + 1}} + 1} \right)}}\\
    = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\left( {\sqrt[5]{{5x + 1}}} \right)}^4} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^3} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^2} + \sqrt[5]{{5x + 1}} + 1}}\\
    = \dfrac{1}{{1 + 1 + 1 + 1 + 1}}\\
    = \dfrac{1}{4}\\
   \mathop {\lim }\limits_{x \to 8} \dfrac{{\sqrt {9 + 2x}  – 5}}{{\sqrt[3]{x} – 2}}\\
    = \mathop {\lim }\limits_{x \to 8} \dfrac{{\left( {9 + 2x – 25} \right)\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\left( {x – 8} \right)\left( {\sqrt {9 + 2x}  + 5} \right)}}\\
    = \mathop {\lim }\limits_{x \to 8} \dfrac{{2.\left( {x – 8} \right)\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\left( {x – 8} \right)\left( {\sqrt {9 + 2x}  + 5} \right)}}\\
    = \mathop {\lim }\limits_{x \to 8} \dfrac{{2\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\sqrt {9 + 2x}  + 5}}\\
    = \dfrac{{2.\left( {\sqrt[3]{{{8^2}}} + 2.\sqrt[3]{8} + 4} \right)}}{{\sqrt {9 + 2.8}  + 5}}\\
    = \dfrac{{2.\left( {4 + 4 + 4} \right)}}{{5 + 5}}\\
    = \dfrac{{12}}{5}
   \end{array}$

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