1) Tính các giới hạn:a) \(\lim \frac{{3{n^2} + 1}}{{{n^2} - 2}}.

1) a) \(\lim \frac{{3{n^2} + 1}}{{{n^2} - 2}} = \lim \frac{{{n^2}\left( {3 + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {1 - \frac{2}{{{n^2}}}} \right)}} = \lim \frac{{3 + \frac{1}{{{n^2}}}}}{{1 - \frac{2}{{{n^2}}}}} = \frac{3}{1} = 3\)

b) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {{x^2} + 5}  - 3}}{{2 - x}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt {{x^2} + 5}  - 3} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}{{\left( {2 - x} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{\left( {2 - x} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {2 - x} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x + 2} \right)}}{{\sqrt {{x^2} + 5}  + 3}} = \frac{{ - \left( {2 + 2} \right)}}{{\sqrt {{2^2} + 5}  + 3}} = \frac{{ - 2}}{3}
\end{array}\)

2) 

TXĐ: D = R

Ta có: \(f(-1)=-m-2m^2\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ + }} \frac{{{x^2} - x - 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - {1^ + }} \frac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - {1^ + }} \left( {x - 2} \right) =  - 3\\
\mathop {\lim }\limits_{x \to  - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ - }} \left( {mx - 2{m^2}} \right) =  - m - 2{m^2}
\end{array}\)

Hàm số đã cho liên tục tại điểm x = - 1 khi và chỉ khi 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ - }} f\left( x \right) = f\left( { - 1} \right) \Leftrightarrow  - m - 2{m^2} =  - 3 \Leftrightarrow 2{m^2} + m - 3 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = \frac{{ - 3}}{2}
\end{array} \right.
\end{array}\)

Vậy các giá trị cần tìm của m là \(m \in \left\{ {1; - \frac{3}{2}} \right\}\)