Giải thích các bước giải:

\(\begin{array}{l}
2.\\
c,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3}  + {x^2} + x - 4}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {x + 3}  - 2} \right) + \left( {{x^2} + x - 2} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {x + 3} \right) - {2^2}}}{{\sqrt {x + 3}  + 2}} + \left( {x - 1} \right)\left( {x + 2} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{x - 1}}{{\sqrt {x + 3}  + 2}} + \left( {x - 1} \right)\left( {x + 2} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{1}{{\sqrt {x + 3}  + 2}} + x + 2} \right]\\
 = \frac{1}{{\sqrt {1 + 3}  + 2}} + 1 + 2\\
 = \frac{1}{4} + 3 = \frac{{13}}{4}\\
d,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - 1}  - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x - 1}  - x} \right) + \left( {x - \sqrt[3]{{3{x^2} - 3x + 1}}} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {2x - 1} \right) - {x^2}}}{{\sqrt {2x - 1}  + x}} + \frac{{{x^3} - \left( {3{x^2} - 3x + 1} \right)}}{{{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}}}}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{ - {{\left( {x - 1} \right)}^2}}}{{\sqrt {2x - 1}  + x}} + \frac{{{{\left( {x - 1} \right)}^3}}}{{{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}}}}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{ - 1}}{{\sqrt {2x - 1}  + 1}} + \frac{{x - 1}}{{{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}}}} \right]\\
 = \frac{{ - 1}}{{\sqrt {2.1 - 1}  + 1}} + \frac{{1 - 1}}{{{1^2} + 1.\sqrt[3]{{{{3.1}^2} - 3.1 + 1}} + {{\sqrt[3]{{{{3.1}^2} - 3.1 + 1}}}^2}}}\\
 = \frac{{ - 1}}{2}
\end{array}\)

Bài 3:

Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {4 - 3{x^2}} \right) = 4 - 3.{\left( { - 2} \right)^2} =  - 8\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} {x^3} = {\left( { - 2} \right)^3} =  - 8\\
 \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)
\end{array}\)

Suy ra hàm số đã cho liên tục tại điểm \(x =  - 2\)