Đáp án:
$\begin{array}{l}
1)\lim \dfrac{{{x^2} - 1}}{{x - 1}} = \lim \left( {x + 1} \right) = \infty \\
2)a)y = {x^4} + 2{x^2} + 2\\
\Leftrightarrow y' = 4{x^3} + 4x\\
b)y = \dfrac{{3x + 2}}{{x + 1}}\\
\Leftrightarrow y' = \dfrac{{3.\left( {x + 1} \right) - 3x - 2}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\\
3)y = {x^3} - 3{x^2} + 2\\
\Leftrightarrow y' = 3{x^2} - 6x\\
+ Tại:M\left( {1;0} \right)\\
\Leftrightarrow PTTT:y = {y_1}'\left( {x - 1} \right) + 0\\
\Leftrightarrow y = \left( {{{3.1}^2} - 6.1} \right).\left( {x - 1} \right)\\
\Leftrightarrow y = - 3x + 3
\end{array}$