Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 3x + 2}}{{{x^3} - 4{x^2} + 2x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {{x^3} - {x^2}} \right) - \left( {3{x^2} - 3x} \right) - \left( {x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{{x^2}\left( {x - 1} \right) - 3x\left( {x - 1} \right) - \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} - 3x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} - 3x - 1}} = \frac{{1 - 2}}{{{1^2} - 3.1 - 1}} = \frac{1}{3}\\
f\left( 1 \right) = {m^2}.1 - 100 = {m^2} - 100
\end{array}\)

Hàm số đã cho liên tục tại \(x = 1\) khi và chỉ khi:

\(\mathop {\lim }\limits_{x \to 1}  = f\left( 1 \right) \Leftrightarrow {m^2} - 100 = \frac{1}{3} \Rightarrow {m^2} = \frac{{301}}{3} \Rightarrow m =  \pm \frac{{\sqrt {903} }}{3}\)

\(\begin{array}{l}
b,\\
\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ + }} \frac{{\sqrt {x - 4}  - \sqrt {x + 4}  + 2}}{{x - 5}}\\
 = \mathop {\lim }\limits_{x \to {5^ + }} \left[ {\frac{{\sqrt {x - 4}  - 1}}{{x - 5}} + \frac{{3 - \sqrt {x + 4} }}{{x - 5}}} \right]\\
 = \mathop {\lim }\limits_{x \to {5^ + }} \left[ {\frac{{\left( {\sqrt {x - 4}  - 1} \right)\left( {\sqrt {x - 4}  + 1} \right)}}{{\left( {x - 5} \right)\left( {\sqrt {x - 4}  + 1} \right)}} + \frac{{\left( {3 - \sqrt {x + 4} } \right)\left( {3 + \sqrt {x + 4} } \right)}}{{\left( {x - 5} \right)\left( {3 + \sqrt {x + 4} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to {5^ + }} \left[ {\frac{{\left( {x - 4} \right) - {1^2}}}{{\left( {x - 5} \right)\left( {\sqrt {x - 4}  + 1} \right)}} + \frac{{{3^2} - \left( {x + 4} \right)}}{{\left( {x - 5} \right)\left( {3 + \sqrt {x + 4} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to {5^ + }} \left[ {\frac{{x - 5}}{{\left( {x - 5} \right)\left( {\sqrt {x - 4}  + 1} \right)}} + \frac{{5 - x}}{{\left( {x - 5} \right)\left( {3 + \sqrt {x + 4} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to {5^ + }} \left[ {\frac{1}{{\sqrt {x - 4}  + 1}} - \frac{1}{{3 + \sqrt {x + 4} }}} \right]\\
 = \frac{1}{{\sqrt {5 - 4}  + 1}} - \frac{1}{{3 + \sqrt {5 + 4} }} = \frac{1}{3}\\
\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ - }} \left( {2a{x^2} - 3x + 1} \right) = 2.a{.5^2} - 3.5 + 1 = 50a - 14
\end{array}\)

Hàm số đã cho liên tục tại \(x = 5\) khi và chỉ khi:

\(\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) \Leftrightarrow 50a - 14 = \frac{1}{3} \Rightarrow a = \frac{{43}}{{150}}\)