Đáp án:
\[A = 23\frac{{23}}{{36}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right)\\
= \mathop {\lim }\limits_{x \to {3^ - }} \left( {A + \frac{{\sqrt {x + 6} + 2x - 9}}{{{x^3} - 4{x^2} + 3x}}} \right)\\
= A + \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\left( {\sqrt {x + 6} - 3} \right) + \left( {2x - 6} \right)}}{{x\left( {x - 1} \right)\left( {x - 3} \right)}}\\
= A + \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\frac{{x + 6 - 9}}{{\sqrt {x + 6} + 3}} + 2.\left( {x - 3} \right)}}{{x\left( {x - 1} \right)\left( {x - 3} \right)}}\\
= A + \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\frac{1}{{\sqrt {x + 6} + 3}} + 2}}{{x\left( {x - 1} \right)}}\\
= A + \frac{{\frac{1}{{\sqrt {3 + 6} + 3}} + 2}}{{3.\left( {3 - 1} \right)}}\\
= A + \frac{{13}}{{36}}\\
\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {3{x^2} - 3} \right) = {3.3^2} - 3 = 24
\end{array}\)
Hàm số đã cho có giới hạn tại x=3 khi và khi khi:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)\\
\Leftrightarrow A + \frac{{13}}{{36}} = 24\\
\Leftrightarrow A = 23\frac{{23}}{{36}}
\end{array}\)
