Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\lim \frac{{4{n^2} - 3n}}{{8{n^2} + 7}} = \lim \frac{{4 - \frac{3}{n}}}{{8 + \frac{7}{{{n^2}}}}} = \frac{{4 - 0}}{{8 + 0}} = \frac{1}{2}\\
b,\\
\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x + 4}}{{x - 3}}\\
\mathop {\lim }\limits_{x \to {3^ + }} \left( {x + 4} \right) = 3 + 4 = 7\\
\left. \begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} \left( {x - 3} \right) = 3 - 3 = 0\\
x \to {3^ + } \Rightarrow x > 3 \Rightarrow x - 3 > 0
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \left( {x - 3} \right) = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \frac{{x + 4}}{{x - 3}} = + \infty \\
2,\\
\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + 3x + 2}}{{x + 2}} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{x + 2}} = \mathop {\lim }\limits_{x \to - 2} \left( {x + 1} \right) = - 2 + 1 = - 1\\
f\left( { - 2} \right) = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - 2} f\left( x \right) \ne f\left( { - 2} \right)
\end{array}\)
Suy ra hàm số đã cho không liên tục tại \(x = - 2\)
\(\begin{array}{l}
3,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {x + \sqrt {{x^2} + mx + 2} } \right) = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {x + \sqrt {{x^2} + mx + 2} } \right)\left( {x - \sqrt {{x^2} + mx + 2} } \right)}}{{x - \sqrt {{x^2} + mx + 2} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - \left( {{x^2} + mx + 2} \right)}}{{x - \sqrt {{x^2}\left( {1 + \frac{m}{x} + \frac{2}{{{x^2}}}} \right)} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{ - mx - 2}}{{x - \left| x \right|.\sqrt {1 + \frac{m}{x} + \frac{2}{{{x^2}}}} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{ - mx - 2}}{{x + x\sqrt {1 + \frac{m}{x} + \frac{2}{{{x^2}}}} }} = 2\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{ - m - \frac{2}{x}}}{{1 + \sqrt {1 + \frac{m}{x} + \frac{2}{{{x^2}}}} }} = 2\\
\Leftrightarrow \frac{{ - m}}{{1 + \sqrt 1 }} = 2\\
\Leftrightarrow m = - 4
\end{array}\)
