Đáp án:
\(\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {2x + 1} + 3{x^2} - 13x + 1}}{{{x^2} - 5x + 4}} = \frac{{34}}{9}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {2x + 1} + 3{x^2} - 13x + 1}}{{{x^2} - 5x + 4}}\\
= \mathop {\lim }\limits_{x \to 4} \left[ {\frac{{\sqrt {2x + 1} - 3}}{{{x^2} - 5x + 4}} + \frac{{3{x^2} - 13x + 4}}{{{x^2} - 5x + 4}}} \right]\\
= \mathop {\lim }\limits_{x \to 4} \left[ {\frac{{\left( {\sqrt {2x + 1} - 3} \right)\left( {\sqrt {2x + 1} + 3} \right)}}{{\left( {{x^2} - 5x + 4} \right)\left( {\sqrt {2x + 1} + 3} \right)}} + \frac{{\left( {3{x^2} - 12x} \right) - \left( {x - 4} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 4} \left[ {\frac{{\left( {2x - 1} \right) - {3^2}}}{{\left( {x - 1} \right)\left( {x - 4} \right)\left( {\sqrt {2x + 1} + 3} \right)}} + \frac{{\left( {x - 4} \right)\left( {3x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 4} \left[ {\frac{{2\left( {x - 4} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)\left( {\sqrt {2x + 1} + 3} \right)}} + \frac{{3x - 1}}{{x - 1}}} \right]\\
= \mathop {\lim }\limits_{x \to 4} \left[ {\frac{2}{{\left( {x - 1} \right)\left( {\sqrt {2x + 1} + 3} \right)}} + \frac{{3x - 1}}{{x - 1}}} \right]\\
= \frac{2}{{\left( {4 - 1} \right).\left( {\sqrt {2.4 + 1} + 3} \right)}} + \frac{{3.4 - 1}}{{4 - 1}}\\
= \frac{{34}}{9}
\end{array}\)
