mathop lim limitsx to 1 căn 1 + 3x + căn [3]1 - 9x( x - 1 )^2

Lời giải của Tự Học 365

Giải chi tiết:

\(\begin{array}{l}
= \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 + 3x} - \left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + \sqrt[3]{{1 - 9x}} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 + 3x} - \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}}{{{{\left( {x - 1} \right)}^2}}} + \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{1 - 9x}} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {\sqrt {1 + 3x} - \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]\left[ {\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]}}{{{{\left( {x - 1} \right)}^2}\left[ {\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {\sqrt[3]{{1 - 9x}} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]\left[ {{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}} \right]}}{{{{\left( {x - 1} \right)}^2}\left[ {{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}} \right]}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{1 + 3x - {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left[ {\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]}} + \mathop {\lim }\limits_{x \to 1} \frac{{1 - 9x + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^3}}}{{{{\left( {x - 1} \right)}^2}\left[ {{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}} \right]}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{1 + 3x - \frac{1}{{16}}\left( {9{x^2} + 30x + 25} \right)}}{{{{\left( {x - 1} \right)}^2}\left[ {\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]}} + \mathop {\lim }\limits_{x \to 1} \frac{{1 - 9x + \frac{1}{{64}}\left( {27{x^3} + 135{x^2} + 225x + 125} \right)}}{{{{\left( {x - 1} \right)}^2}\left[ {{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}} \right]}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{ - \frac{9}{{16}}{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left[ {\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)} \right]}} + \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{27}}{{64}}{{\left( {x - 1} \right)}^2}\left( {x + 7} \right)}}{{{{\left( {x - 1} \right)}^2}\left[ {{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}} \right]}}\\
= - \frac{9}{{16}}\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {1 + 3x} + \left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}} + \frac{{27}}{{64}}\mathop {\lim }\limits_{x \to 1} \frac{{x + 7}}{{{{\sqrt[3]{{1 - 9x}}}^2} - \sqrt[3]{{1 - 9x}}\left( {\frac{{3x}}{4} + \frac{5}{4}} \right) + {{\left( {\frac{{3x}}{4} + \frac{5}{4}} \right)}^2}}}\\
= - \frac{9}{{16}}.\frac{1}{{2 + 2}} + \frac{{27}}{{64}}.\frac{8}{{{{\left( { - 2} \right)}^2} + 2.2 + {2^2}}} = \frac{9}{{64}}
\end{array}\)