Đáp án: 10, a Có, b không, c có

               11 a, $A=0$, b, $A=\dfrac{877}{36}$

Lời giải:

10. 

\(\begin{array}{l}
a)\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} - 3x + 2}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{x - 2}}{{x + 1}} =  - \dfrac{1}{2}\\
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{ - x}}{2} = \dfrac{{ - 1}}{2}\\
 \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\text{ nên hàm số có giới hạn tại }x_o=1\\
b)\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {1 - 2x} \right) =  - 3\\
\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{4 - {x^2}}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{\left( {2 - x} \right)\left( {x + 2} \right)}}{{x - 2}}\\
 = \mathop {\lim }\limits_{x \to {2^ - }}  - \left( {x + 2} \right) =  - 4\\
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\text{ nên hàm số không có giới hạn tại }x_o=2\\
c)\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {x + 1}  - 1}}{{\sqrt[3]{{1 + x}} - 1}}\\
 = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left( {x + 1 - 1} \right)\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}}{{\left( {1 + x - 1} \right)\left( {\sqrt {x + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1}}{{\sqrt {x + 1}  + 1}} = \dfrac{3}{2}
\end{array}\)

Nên hàm số có giới hạn tại $x_o=0$

11

a) $f(x)=\left\{\begin{array}{I}\dfrac{x^2-1}{x-1}(x>1)\\Ax+2(x\le1)\end{array}\right.$ với $x_o=1$

$\lim\limits_{x\to{1^+}}f(x)=\lim\limits_{x\to{1^+}}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to{1^+}}\dfrac{(x-1)(x+1)}{x-1}=\lim\limits_{x\to{1^+}}x+1=2$

$\lim\limits_{x\to{1^-}}f(x)=\lim\limits_{x\to{1^-}}Ax+2=-A+2$

Để $f(x)$ có giới hạn tại $x_o=1$ thì

$\lim\limits_{x\to{1^-}}f(x)=\lim\limits_{x\to{1^-}}$

$\Rightarrow -A+2=2\Rightarrow A=0$

b) $\lim\limits_{x\to{3^-}}f(x)=\lim\limits_{x\to{3^-}}A+\dfrac{\sqrt{x+6}+2x-9}{x^2-4x^2+3x}$

$=\lim\limits_{x\to{3^-}}A+\dfrac{(\sqrt{x+6}-3)+2x-6}{x(x-1)(x-3)}$

$=\lim\limits_{x\to{3^-}}A+\dfrac{\sqrt{x+6}-3}{x(x-1)(x-3)}+\dfrac{2x-6}{x(x-1)(x-3)}$

$=\lim\limits_{x\to{3^-}}A+\dfrac{x+6-9}{x(x-1)(x-3)(\sqrt{x+6}+3)}+\dfrac{2(x-3)}{x(x-1)(x-3)}$

$=\lim\limits_{x\to{3^-}}A+\dfrac{1}{x(x-1)(\sqrt{x+6}+3)}+\dfrac{2}{x(x-1)}$

$=A+\dfrac{1}{36}+\dfrac{1}{3}=A+\dfrac{13}{36}$

$\lim\limits_{x\to{3^+}}f(x)=\lim\limits_{x\to{3^+}}3x^2-2=25$

Để $f(x)$ có giới hạn tại $x_o=3$ thì

$=\lim\limits_{x\to{3^-}}f(x)=\lim\limits_{x\to{3^+}}$

$\Rightarrow A+\dfrac{13}{36}=25\Rightarrow A=\dfrac{887}{36}$