Đáp án:
$a =-\dfrac{47}{12}$
Giải thích các bước giải:
\(\begin{array}{l}
\quad f(x) = \begin{cases}2x + a + 3\qquad \qquad \quad khi\quad x =0\\\dfrac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}\quad khi \quad x \ne 0 \end{cases}\\
\text{Ta có:}\\
+)\quad \lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\dfrac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}\\
\to \lim\limits_{x\to 0}f(x) =\lim\limits_{x\to 0}\dfrac{2\sqrt{1-x} - 2 + 2 - \sqrt[3]{8-x}}{x}\\
\to \lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\dfrac{2(\sqrt{1-x} - 1)}{x} + \lim\limits_{x\to 0}\dfrac{2-\sqrt[3]{8-x}}{x}\\
\to \lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\dfrac{-2}{\sqrt{1-x} + 1} + \lim\limits_{x\to 0}\dfrac{1}{4 + 2\sqrt[3]{8-x} + \sqrt[3]{(8-x)^2}}\\
\to \lim\limits_{x\to 0}f(x) = \dfrac{-2}{\sqrt{1 - 0} + 1} + \dfrac{1}{4 + 2\sqrt[3]{8-0} + \sqrt[3]{(8-0)^2}}\\
\to \lim\limits_{x\to 0}f(x) = -\dfrac{11}{12}\\
+)\quad f(0) = 2.0 + a + 3\\
\to f(0) = a + 3\\
\text{Hàm số liên tục tại $x=0$ khi và chỉ khi}\\
\quad \lim\limits_{x\to 0}f(x) = f(0)\\
\Leftrightarrow a + 3 = -\dfrac{11}{12}\\
\Leftrightarrow a = -\dfrac{47}{12}
\end{array}\)
