Tính giới hạn \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + x + 1} \right).

* Hướng dẫn giải

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + x + 1} \right) = 7.\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 9}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} (x + 3) = 6.\)

c) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3}  + {x^2} + x - 4}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{\sqrt {x + 3}  - 2}}{{x - 1}} + \frac{{{x^2} + x - 2}}{{x - 1}}} \right)\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}} + x + 2} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{{\left( {\sqrt {x + 3}  + 2} \right)}} + x + 2} \right) = \frac{{13}}{4}.
\end{array}\)

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2{\rm{x}} - 1}  - \sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}}}}{{{{\left( {x - 1} \right)}^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2{\rm{x}} - 1}  - x + x - \sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}}}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{\sqrt {2{\rm{x}} - 1}  - x}}{{{{\left( {x - 1} \right)}^2}}} + \frac{{x - \sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}}}}{{{{\left( {x - 1} \right)}^2}}}} \right)\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{ - {{(x - 1)}^2}}}{{{{\left( {x - 1} \right)}^2}\left( {\sqrt {2{\rm{x}} - 1}  + x} \right)}} + \frac{{{{(x - 1)}^3}}}{{{{\left( {x - 1} \right)}^2}({x^2} + x\sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}}} \right)}^2})}}} \right)\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{ - 1}}{{\sqrt {2{\rm{x}} - 1}  + x}} + \frac{{(x - 1)}}{{{x^2} + x\sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{3{{\rm{x}}^2} - 3{\rm{x}} + 1}}} \right)}^2}}}} \right) =  - \frac{1}{2}.
\end{array}\)