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Đáp án:

 C2: m=4

Giải thích các bước giải:

\(\begin{array}{l}
C1:\\
a)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {1 + 3{n^2}} }}{{1 - 5n}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {\dfrac{1}{{{n^2}}} + 3} }}{{\dfrac{1}{n} - 5}}\\
 = \dfrac{{\sqrt 3 }}{{ - 5}}\\
b)\mathop {\lim }\limits_{x \to 1} \dfrac{{2{x^2} - 3x + 1}}{{\left( {1 - x} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {2x - 1} \right)}}{{\left( {1 - x} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - 1}}{{ - \left( {x + 1} \right)}} = \dfrac{{2.1 - 1}}{{ - \left( {1 + 1} \right)}} =  - \dfrac{1}{2}\\
C2:\\
Xét:\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 2x - 3}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 3} \right)\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 3} \right) = 4\\
f\left( 1 \right) = m
\end{array}\)

Để hàm số liên tục tại \({x_0} = 1\)

\(\begin{array}{l}
 \to f\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right)\\
 \to m = 4
\end{array}\)