Đáp án: a=-3
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\\
= \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {1 - x} - \sqrt {1 + x} }}{x}\\
= \mathop {\lim }\limits_{x \to {0^ - }} \frac{{1 - x - 1 - x}}{{x\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2}}{{\sqrt {1 - x} + \sqrt {1 + x} }}\\
= \frac{{ - 2}}{{\sqrt {1 - 0} + \sqrt {1 + 0} }}\\
= - 1\\
\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {a + \frac{{4 - x}}{{x + 2}}} \right)\\
= a + \frac{{4 - 0}}{{0 + 2}}\\
= a + 2
\end{array}$
Để hs liên tục tại x=0 thì:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\\
\Rightarrow - 1 = a + 2\\
\Rightarrow a = - 3
\end{array}$