Đáp án:
\(\begin{array}{l}
1b)\quad -\infty\\
2)\quad a = \dfrac73
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1b)\quad \lim\limits_{x\to -\infty}\left(3x^3 - 2x + 1\right)\\
= \lim\limits_{x\to -\infty}\left[x^3\left(3 - \dfrac{2}{x^2} + \dfrac{1}{x^3}\right)\right]\\
= \lim\limits_{x\to -\infty}x^3\cdot \lim\limits_{x\to -\infty}\left(3 - \dfrac{2}{x^2} + \dfrac{1}{x^3}\right)\\
= -\infty \cdot 3\\
= -\infty\\
2)\quad f(x) = \begin{cases}\dfrac{x^2 - 9}{x-3}\quad khi\quad x\ne 3\\
ax - 1\quad \ khi\quad x = 3\end{cases}\\
\text{Ta có:}\\
+)\quad f(3) = 3a - 1\\
+)\quad \lim\limits_{x\to 3}f(x)\\
= \lim\limits_{x\to 3}\dfrac{x^2 - 9}{x-3}\\
= \lim\limits_{x\to 3}\dfrac{(x-3)(x+3)}{x-3}\\
= \lim\limits_{x\to 3}(x+3)\\
= 3 + 3 = 6\\
\text{Hàm số liên tục tại $x=3$}\\
\Leftrightarrow f(3) = \lim\limits_{x\to 3}f(x)\\
\Leftrightarrow 3a - 1 = 6\\
\Leftrightarrow a = \dfrac73
\end{array}\)
