Lim x->3 (x^3 - 4x^2+4x-3)/(x^2-3x) cho điểm luôn

Đáp án:

\[\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - 4{x^2} + 4x - 3}}{{{x^2} - 3x}} = \frac{7}{3}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - 4{x^2} + 4x - 3}}{{{x^2} - 3x}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {{x^3} - 3{x^2}} \right) - \left( {{x^2} - 3x} \right) + \left( {x - 3} \right)}}{{x\left( {x - 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2}\left( {x - 3} \right) - x\left( {x - 3} \right) + \left( {x - 3} \right)}}{{x\left( {x - 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {{x^2} - x + 1} \right)}}{{x\left( {x - 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - x + 1}}{x}\\
 = \frac{{{3^2} - 3 + 1}}{3} = \frac{7}{3}
\end{array}\)